

























Class._ : ' c ■ 

Book_ 


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■>—««-— 


Gojjyjigfit N°. 


COFfluGHT DEPOSIT. 












THERMODYNAMICS, ABRIDGED 








THERMODYNAMICS, ABRIDGED 

A MANUAL AND TEXT BOOK FOR MIDSHIPMEN 
AT THE UNITED STATES NAVAL ACADEMY 


BY 


WILLIAM D. ENNIS, M.E. 


MEM. AM. SOC. M.E., FELLOW A. A. A. 8., CONSULTING ENGINEER: VICE PRESIDENT OF THE 
TECHNICAL ADVISORY CORPORATION OF NEW YORK! FORMERLY PROFESSOR OF 
MECHANICAL AND MARINE ENGINEERING IN THE POST GRADUATE 
DEPARTMENT OF THE UNITED STATES NAVAL ACADEMY 


78 ILLUSTRATIONS 



V £ 

* 9 ■< 

> ** 

NEW YORK 

D. VAN NOSTRAND COMPANY 

Eight Warren Street 

1920 




















Copyright, 1920, by 

D. Van Nostrand Company 



DEC 31 1920 


PRESS OF 

THE NEW ERA PRINTING COMPANY 
LANCASTER, PA. 


2>CI. A604850 


o 


PREFACE 

(WHICH STUDENTS SHOULD READ) 




This brief statement of fundamental principles has been 
prepared especially for the use of midshipmen in the United 
States Naval Academy. The work was undertaken at the 
request of Commander J. 0. Richardson, U. S. N., Head of the 
Department of Marine Engineering and Naval Construction, 
and has been carried on in close cooperation with officers of that 
Department, particularly Comdr. W. L. Friedell, U. S. N., 
Lt. Comdr. T. W. Johnson, Professor of Mathematics, U. S. N., 
and Lt. Comdr. H. W. Boynton, U. S. N. 

Thermodynamics is difficult, but worth while. To some 
extent, it has been simplified by planning the problems for easy 
solution. The table preceding Chapter II will be found useful 
for exponential expressions. The solution of many problems 
is necessary in order that a real grasp of the subject may be 
attained. All problems should be solved with the slide rule. 
This implies that answers will be absolutely reliable only with 
respect to two significant figures, the third figure being estimated. 
An error which may be as high as 1 per cent, is therefore allow¬ 
able. The answers given have been obtained by slide rule, 
and are subject to this error. Other errors may occasionally be 
found during a first year’s use of the book. The student’s 
answer may be right, therefore, even when it disagrees with the 
answer in the book. 

One per cent, accuracy is good enough for almost all practical 
engineering calculations. We rarely know the strength of a 
material with any greater exactness. Even the dimensions of 
structural parts are subject to a comparable error. Good 
engineering is largely a matter of not straining at gnats and 
swallowing camels. 

If there is a “royal road” to any kind of learning, thermo¬ 
dynamics will be found to be the royal road to real compre¬ 
hension of steam machinery. 


v 






















NUMERICAL CONSTANTS AND COMMONER 

SYMBOLS 


1.34 hp. = 1 kw. 

2.3026 = log e x -7- logio x. 

5.403 = 778 144. 

32.17 = acceleration of gravity. 
42.42 B.t.u. per min. = 1 hp. 


746 watts = 1 hp. 

778 ft. lb. = 1 B.t.u. 

1544 = PVm -E wT, Art. 12. 
2545 B.t.u. per hr. = 1 hp. 

33 000 ft. lb. per min. = 1 hp. 


1 980 000 ft. lb. per hr. = 1 hp. 


B.t.u. = British thermal unit(s): Art. 3. 
c = clearance: Arts. 7, 39. 

C = compression ratio: Arts. 38, 149. 

D = displacement: cu. ft: Arts. 7, 42. 
e = efficiency: Arts. 34, 80. 

= base of Napierian system of logarithms. 
e v — volumetric efficiency: Art. 42. 

E = T + I = internal energy of vapor: Art. 63. 

F. = Fahrenheit. 

F = factor of evaporation: Art. 73. 

/ = diagram factor: Art. 104. 

g = 32.17 = acceleration of gravity. 

h = heat of 1 lb. of boiling liquid: Art. 63. 

H = total heat of 1 lb. of dry vapor: Art. 65. 

H' = total heat of 1 lb. of superheated vapor: Art. 83. 

H, H = heat received by a substance: Art. 2. 

2/7 = net amount of heat received or rejected in a cycle: Art. 21. 
hp. = horse power, 
ihp., Ihp. = indicated horse power. 

I = disgregation work: Art. 2. 

7 0 = ice melting effect per ihp.-hr.: Art. 59. 
kw. = kilowatt (s). 

k = specific heat at constant pressure: Art. 14. 

I = specific heat at constant volume: Art. 14. 

L = stroke of an engine, ft.: Art. 7. 

= latent heat of vaporization: Art. 64. 
log = logarithm to the base 10. 
loge = logarithm to the base e = 2.3026 log. 
m = molecular weight: Art. 12. 
n == poly tropic exponent: Art. 24. 


Vll 



viii Numerical Constants and Commoner Symbols 

n = entropy or change of entropy: Art. 68. 
n w = entropy of 1 lb. of liquid: Art. 71. 
n e = entropy of vaporization: Art. 71. 
n 8 = total entropy of dry vapor: Art. 71. 
n' — total entropy of superheated vapor: Art. 83. 

N = r.p.m.: Art. 7. 

p = absolute pressure, lb. per sq. in. (= gauge pressure + 14.696). 

P = absolute pressure, lb. per sq. ft. = 1.44 p. 
p m = mean effective pressure: Arts. 7, 39. 
p 0 = intercooler pressure: Art. 45. 

Q = the quantity of heat in 1 lb. of a vapor: Art. 87. 

R = PV -T- wT: Art. 12. 

r = internal latent heat of vaporization: Art. 65. 

= ratio of expansion: Arts. 106, 162. 
r.p.m. = revolutions per minute, 
s = specific heat: Arts. 3, 27. 
t = temperature Fahrenheit. 

T = t + 460 = absolute temperature: Art. 11. 

T = temperature effect of heat: Art. 2. 
t' = temperature of superheated vapor: Art. 83. 
u, U = velocity, ft. per sec.: Arts. 56, 133. 
v = volume of 1 lb. of a substance. 
v' = volume of 1 lb. of a superheated vapor: Art. 83. 
v w = volume of 1 lb. of liquid: Art. 65. 

V = volume of w lb. of a substance. 
w = weight, lb. 

W, W = external mechanical work done: Arts. 2, 26. 

STF = net external work done in a cycle: Art. 21. 
x = dryness fraction: Art. 74. 

y — k -7- l = ratio of specific heats (Art. 17): also adiabatic exponent 
(Art. 28). 


CONTENTS 

Numerical Constants and Commoner Symbols. vii 

Chapter I: Preliminary—General Principles. Values 

of Exponentials in Chapters II and III. . 1 

Chapter II: Permanent Gases. Laws and Processes... 13 

Chapter III: Air Engines, Air Compressors, Air Refrig¬ 
eration . 46 

Chapter IV: Steam and Other Vapors. 76 

Chapter V: Efficiency and Power of Steam Engines. . . 118 

Chapter VI: Vapor Refrigeration. 138 

Chapter VII: Heat Transfer Appliances. 153 

Chapter VIII: Fluid Flow and the Steam Turbine. 171 

Chapter IX: Internal Combustion Engines. 203 


ix 































































































































































































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THERMODYNAMICS, ABRIDGED 


CHAPTER I 

PRELIMINARY—GENERAL PRINCIPLES 

1. Subject-Matter. Thermodynamics treats of heat as a 
source of motive power. All artificial motive powers (not 
excluding windmills, tide motors and muscular effort) originate 
from heat. Heat is not matter, but energy. It is measurable 
in foot-pounds. Heat may be regarded as due to molecular 
motion. Properly speaking, Thermodynamics is not concerned 
with phenomena of heat transmission, which belong in the field 
of genera] physics. Neither does it traverse the domain of 
thermochemistry, which deals with heats of formation without 
relation to mechanical effects. Thus our subject is from one 
standpoint merely a subdivision of the subject of heat, which 
latter constitutes one of the departments of physics. More 
broadly viewed, however, Thermodynamics deals with molecular 
mechanics while general physics (including practical astronomy) 
is concerned with mass mechanics. 

Questions. How does muscular energy originate from heat? 
Show that the action of a windmill is due to heat. Show that 
the force which propels a steamship is derived from the heat 
of the sun. 

2. Effects of Heat. Thermodynamics ignores electrical and 
chemical effects and (for the present) considers only three results 
following the application of heat to a substance: 

(1) Increase of Temperature. This is the most familiar effect. 
If heat results from molecular motion, temperature may be 
regarded as due to molecular velocity. The molecules are 

1 



2 


Thermodynamics, Abridged 


incessantly vibrating. The faster they move, the higher is the 
temperature. Temperature is measured by the thermometer 
and may be expressed in Fahrenheit or Centigrade units. Ice 
melts, normally, at 0° C. or 32° F.; water boils under standard 
conditions at atmospheric pressure at 100° C. or 212° F. A range 
of 100° C. means the same thing as a range of 180° F. Hence 
1 degree range of Centigrade temperature = 9/5 degree range of 
Fahrenheit temperature, and 

T f = 9/5 T c + 32, (1) 

T c = 5/9 (T f - 32). (2) 

Note that the Fahrenheit freezing point, 32°, enters into both 
equations. 

(2) Disgregation Work. The molecules of fluids exert forces 
upon one another. When heat is supplied, the molecules may 
take up new positions, i.e., assume a new configuration. This 
requires that energy be expended to destroy the original equi¬ 
librium. Potential energy is stored up in establishing a new 
condition of equilibrium. The expenditure of energy necessary 
for rearranging the molecules is called disgregation work. Dis¬ 
gregation work occurs, for example, when ice is melted by heat. 

(3) External Mechanical Work. In general, bodies expand 
when heated. The work done by reason of such expansion 
against resisting objects is called external work and is the 
effect of heat with which we are primarily concerned. 


The action of heat may then be summed up thus: the heat, H, 
received by the substance is utilized, (a) in raising its temperature, 
(6) in rearranging its molecules, and (c) in performing mechanical 
work. In symbols, 

H = T + I + W, (3) 

where T = heat expended in changing temperature, 

I = heat expended in doing disgregation work, 

W — heat expended in doing external work. 

All terms must be expressed in the same unit, which may be 



Preliminary—General Principles 


3 


either a heat unit or a work unit. Signs may be either + or —. 
Thus if heat is emitted, the sign of H is —. The sign of T 
is — when the temperature falls. The sign of W is — if work 
is consumed, i.e., if work is done on the substance instead of 
by it. When a substance expands, W is +. When it is com¬ 
pressed, W is —. Note that T is not the temperature, but the 
quantity of heat consumed in changing the temperature. It is 
not even the change of temperature. The sign of I is + if 
molecular forces have been overcome, as when a separation of 
mutually attractive molecules is effected. Its sign is — in the 
reverse instance, as when such molecules are brought closer 
together. 

Prob. 1. At what temperature do the Centigrade and Fahren¬ 
heit thermometers read alike? (Ans., — 40°.) 

Prob. 2. What is the (algebraic) amount of work done in 
rearranging the molecules if the heat received is 400, the heat 
represented by a rise of temperature 100 and the mechanical 
work done 120? (Ans., + 180.) 

Prob. 3. If IF = - 20, 1 = + 30, T = + 10, find H. 
(Ans., + 20.) 

Prob. 4. When 1 lb. of water at the boiling point is converted 
into steam, 970.4 units of heat must be supplied. Thqjnechani- 
cal work done by the expansion of water into steam is 72.8 units. 
Compute T and I. (Ans., T = 0, I = + 897.6.) 

3. Definitions and Laws. Two bodies are at the same temper¬ 
ature when there is no tendency toward a transfer of heat from 
one to the other. When a heat-transfer occurs, heat always 
passes from a high-temperature body to one of lower temperature. 
The transfer tends to continue until the two temperatures are 
equal. 

The heat unit (British thermal unit, B.t.u.) is the quantity 
of heat required to raise the temperature of 1 lb. of water 1 F. 
Since this quantity is somewhat variable according to the initial 
temperature, we will use the mean B.t.u., defined as 1/180 of 


4 


Thermodynamics, Abridged 


THAT QUANTITY OF HEAT CONSUMED IN RAISING 1 LB. OF WATER 

from 32° F. to 212° F. 

The calorie is the quantity of heat required to raise 1 kilo¬ 
gram (2.2046 lb.) of water 1° C. (9/5° F.). Its value is 

2.2046 X | = 3.9683 B.t.u. 
o 

The specific heat of a substance is the quantity of heat 
(number of heat units) necessary to increase the temperature 
of 1 lb. of that substance 1° F. By definition, then, the mean 
specific heat of water from 32° F. to 212° F. is 1.0. The specific 
heat of a gas is not an absolute constant (see Art. 14). 

When H B.t.u. are supplied to w lb. of a substance having a 
specific heat s, 

II = ws(T 2 — Ti), (4) 

where T 2 and T i are the final and initial temperatures respec¬ 
tively. 

When several fluids are mixed, some will lose heat and others 
will gain heat. The total loss (including loss of heat by any 
surrounding bodies) must equal the total gain. In other words, 
the algebraic sum of heat losses is zero: 

Loss by first body + loss by second body 

+ loss by third body • • • + radiation “loss” = 0 
w a s a (T a — T m ) + w b s b (T b — T m ) 

+ w c s c (T c — T m ) • • • + H r — 0. (5) 

Here iv a , w b , w c , denote weights of the several fluids in lb., 
s a , s b , s c , their respective specific heats, 

Ha, T b , Tc, their respective initial temperatures, 

T m , their common final temperature, 

H r , the heat contributed by surrounding bodies (which 
will be negative if T m exceeds the surrounding temper¬ 
ature) . 

Prob. 5. How many (mean) B.t.u. are required to heat 4 lb. 
of water from 32° to 212 p ? (Ans., 720.) 


Preliminary—General Principles 


5 


Prob. 6 . If 1 meter — 39.37 in., how many kilogram-meters 
are equivalent to 1 ft. lb.? (Ans. 0.1383.) 

Prob. 7. If the specific heat is 0.10, how many calories are 
consumed in heating 10 lb. 100° F.? (Ans., 25.2.) 

Prob. 8. When 1 lb. of water is heated from 100° to 200°, 
99.97 B.t.u. are expended. What is the mean specific heat of 
water for this range? (Ans., 0.9997.) 

Prob. 9. The specific heat is 0.3, the weight 3f lb. How 
many B.t.u. are consumed if the temperature rises 500° F.? 

(Ans., 500 B.t.u.) 

Prob. 10. A mixture is made of three liquids, the weights 
being 5, 3 and 22 lb., the corresponding specific heats 1.0, 0.3 
and 0.12 and the final temperature 189°. The original tempera¬ 
tures were 200°, 110° and 220°. How much heat was lost by 
radiation during the mixing? (Ans., H R = — 65.7 B.t.u.) 

4. Mechanical Equivalent of Heat. The “ first law ” of 
Thermodynamics is, 

Heat and mechanical energy are mutually 
CONVERTIBLE, IN THE RATIO OF 778 FT. LB. (777.5 MORE 
nearly) to 1 B.t.u. 

This law has been established by direct experiment (expending 
work in agitating a fluid and measuring the heating accomplished) 
and from abundant inference. 1 B.t.u. = 778 ft. lb., just as 
1 yard = 3 ft. We do not always realize 778 ft. lb. of useful 
work from 1 B.t.u., because heat does not equal W, but T + 1 
+ W (Equation 3). The number 778 is called the mechanical 

EQUIVALENT OF HEAT. 

In a heat engine test, each pound weight of steam entered the 
cylinder containing 1125 B.t.u. Leaving the cylinder, each 
pound contained 1000 B.t.u. Then 125 B.t.u. disappeared. 
The engine developed 150 horsepower = 150 X 33 000 or 
4 950 000 ft. lb. per min. It used 55 lb. of steam per min., 
or 55 X 125 = 6875 B.t.u., if we assume the heat which dis¬ 
appeared to have been “ used.” Then 6875 B.t.u. = 4 950 000 


6 


Thermodynamics, Abridged 


ft. lb. or 1 B.t.u. = 720 ft. lb. This is as close an approximation 
to the true value as might be expected from this type of experi¬ 
ment. 

Prob. 11 . How many kilogram-meters are equivalent to 1 
calorie (see Prob. 6). (Ans., 427.) 

Prob. 12. 1 lb. of coal contains 14 000 B.t.u. If this could 

be fully utilized, how much coal would be burned per min. to 
develop 1 hp.? (Ans., 0.00303 lb.) 

Prob. 13a. If a 100 hp. engine operates so as to cause the 
disappearance of 4250 B.t.u. per min., what is the probable value 
of the mechanical equivalent of heat? (Ans., 778.) 

5. Efficiency. Efficiency is in general effect -t- cause or 
output -T- input. In Thermodynamics, efficiency is mechani¬ 
cal work done -r- total heat received. Units must be alike. 
A horse power (hp.) is 33 000 ft. lb. per min. or 1 980 000 ft. 
lb. per hr. It is also 42.42 B.t.u. per min. or 2545 B.t.u. per hr. 
At 100 per cent, plant efficiency, an engine of 1 hp. would con¬ 
sume 2545 B.t.u. (in the coal) per hr. At 10 per cent, efficiency 
it would consume 25 450 B.t.u. The efficiency of a mechanical 
device (a chain hoist, for example) may be increased indefinitely 
by fine fitting, ball bearings, etc., with 100 per cent, as the 
ultimate limit. Efficiencies of heat engines have a much lower 
(though for given conditions, a perfectly definite) limit. The 
actual efficiency of a given engine will always be below this 
limit and will vary with the conditions according to laws which 
Thermodynamics aims to discover. Thermodynamics shows 
how big to make an engine and how efficient it is likely to be, 
and why. 

Prob. 13b. A ship uses 2 lb. of coal per hr. per indicated hp. 
The heat value of the coal is 12 725 B.t.u. per lb. What is 
the efficiency from fuel to cylinder? (Ans., 0.10, or 10 per 
cent.) 

Prob. 14. An oil engine has an efficiency of 0.13, or 13 per 
cent. It develops 100 hp. using oil of 19 000 B.t.u. per lb. 
What weight of oil will it consume per hr.? (Ans., 102 lb.) 


Preliminary—General Principles 


7. 


6. Friction: Traction: Plant Efficiency. Thermodynamics is 
concerned essentially with the conversion of heat in steam to 
work in the cylinder of the engine. The corresponding 
efficiency, E T , is called the thermal efficiency. Mechanical 
friction makes the work at the shaft or “ brake ” less than 
that done in the cylinder. The ratio of shaft hp. to cylinder 
hp. is E m , the mechanical efficiency. The efficiency from 
steam to shaft is E T X E M . The efficiency of a series of con¬ 
versions is always the product of the efficiencies of the individual 
conversions: E = Ei X E% X E s • • •. In a turbine, there is no 
“cylinder” and the only efficiency determinable is the ratio of 
work at the shaft to heat supplied by the steam. 

If the resistance of a ship or railway train or other propelled 
object is R lb., the actual hp. expended in propulsion is 

RV_RVm_ RVk 

550 “ 375 “ 326 ’ 

where V, V M and V K are speeds in ft. per sec., miles per hr. and 
knots per hr., respectively. 

Prob. 15. The shaft hp. is 80, the horse power lost in friction 
is 20 and 2 545 000 B.t.u. are consumed per hr. Find E T 
and E m . (Ans., 0.10, 0.80.) 

Prob. 16. In an electric central station, 2 lb. of coal (14 000 
B.t.u. per lb.) are used per kw. hr. (1 kw. = 1.34 hp.). Effi¬ 
ciencies are as follows: fuel to steam, 0.75: steam to work in 
cylinder, 0.25: shaft to switchboard output, 0.8. Find efficiency 
from cylinder to shaft. (Ans., 0.81.) 

Prob. 17. The battleship Pennsylvania has a resistance of 
254 000 lb. at 20 knots. If E M = 0.80 and the efficiency of the 
propellors is 0.70, what horse power is developed in the cylinders? 

(Ans., 27 700.) 

7. Engine Action. Fig. 1 shows the piston P moving in the 
cylinder C and communicating via the piston rod R with the 
external mechanism. Inlet and outlet passages, I and E, are 
provided with valves, not shown. The piston being in its ex- 

2 


8 


Thermodynamics, Abridged 


treme left-hand position (i.e., as far to the left as the external 
mechanism will permit), the distance a is called the linear 
clearance. The volume then included between the left face 


Extreme right ha not 

position of piston^ 

\ 



of the piston and the closed faces of the valves is the clearance 
volume, v c . The piston now makes one full stroke to the right. 
The length of stroke, or distance moved, is determined by the 
external mechanism. If the whole volume swept through by 
the piston in such a stroke is D, then the clearance (proportion 
of clearance) is c = vJD. If d = piston diameter, L = piston 
stroke, D = (7r/4)d 2 L. The quantity D is called the displace¬ 


ment PER STROKE. 


















Preliminary—General Principles 


9 


Most steam engines have the right-hand end of the cylinder 
closed by a head with inlet and outlet passages. The rod then 
passes through a stuffing box in the head, not shown. The 
engine is then called double-acting, steam being used on both 
sides of the piston. Most gas and oil engines are single-acting. 
In single-acting engines, the right-hand end of the cylinder is 
left open. This gives a better opportunity for cooling the piston 
and piston rod without circulating cold water through them. 
In double-acting engines there is of course a clearance at each 
end of the cylinder. 

The indicator depicts on paper the pressure existing at 
various points (piston positions) during the stroke. The lower 
part of Fig. 1 shows a record of this kind, called an indicator 
Diagram. Ordinates represent pressures on the left-hand side 
of the piston, abscissas represent piston movements. The upper 
line 123 represents the pressures applied while the piston is 
moving to the right. The lower line 341 represents resisting 
pressures during the return (left-hand) stroke. The diagram 
shows what occurs on the left-hand side, only, of the piston. 
It records two strokes. Under constant conditions, subsequent 
records would simply duplicate this one. 

Abscissas (which represent piston movements) also represent 
volumes (to another scale), since the cylinder diameter is the 
same at all points. The volume swept through = area of 
piston X distance piston has moved. The vertical axis (of zero 
volume) falls directly below the inner face of the left-hand cyl¬ 
inder head only when the valve faces are flush with that face. 
This is a condition which rarely if ever occurs. The ov axis is 
located at zero absolute pressure, i.e., 14.696 lb. per sq. in. below 
the pressure of the atmosphere. Obviously c = a -r- b. In a 
double-acting engine, a symmetrically disposed oppositely placed 
diagram represents the action on the right-hand side of the 
piston. 

The average ordinate of the indicator diagram (= area -f- 6, 
with proper consideration of scale) is p m , the mean effective 
pressure, usually taken in lb. per sq. in. It is the average 


10 


Thermodynamics, Abridged 


pressure exerted on one side of the piston during one stroke, the 
pressure during the succeeding stroke being regarded as zero. 
In a double-acting engine, however, there is an approximately 
equal average pressure on the other side of thfe piston during 
this succeeding stroke: hence the average pressure p m is main¬ 
tained continuously. If d = piston diameter in inches, the 
average total pressure on the piston (disregarding the reduction 
of area on one side by the rod) is (7r/4 )d 2 p m lb. If L = stroke 
in ft., A = piston area in sq. in. = (?r/4)d 2 , and the engine 
makes N r.p.m., the pressure p m acts through a distance of L ft. 
per stroke, 2 L ft. per revolution and 2 LN ft. per min., and the 
cylinder horse power is 


ihp. = (7r/4)d 2 p m X 


2 LN 
33 000 


TT'PmLd 2 N 
66 000 


VmAS _ 2 p m ND _ 2 p m LAN 

33 000 “ 12 X 33 000 ” 33 000 ’ W 


where S = piston speed = 2 LN and D = (x/4 )d 2 L X 12. 

The overload capacity of an engine is (maximum power 
— normal power) -5- normal power. If p m is the same at all 
powers, 

N' — N 

overload capacity = — r —, (7) 

where N = normal speed and N ' = maximum speed. If N is 
constant for all powers, 

overload capacity = —-—, (8) 

'Pm 

where p m is the normal value and p m ' the maximum value. In 
most engines, p m varies. In marine engines, both p m and N vary. 

The mean effective pressure fully determines the power of an 
engine of given size and speed. 

Prob. 18. In an engine 10 in. diameter by 20 in. stroke the 
linear clearance is \ in. Find the displacement per stroke in 
cu. in. If the valve faces are flush with the inner face of the 








Preliminary—General Principles 


11 


cylinder head, what is the clearance volume? The proportion of 
clearance? (Ans., 1570.8: 19.635 cu. in.: 0.0125 or lj per cent.) 

Prob. 19. The clearance volume of a 10 by 20 (diameter by 
stroke) in. engine is found to hold 58.905 cu: in. of water. What 
is its clearance? (Ans., 3f per cent.) 

Prob. 20. The indicator diagram from a 10 by 20 in. engine 
is 3.1416 in. long. What displacement is represented by 1 in. 
of diagram length? (Ans., 500 cu. in.) If the clearance of 
this engine is 5 per cent., what will be the dimension a , in Fig. 1? 

(Ans., 0.15708 in.) 

Prob. 21. The area of an indicator diagram is 3 sq. in. and 
its length is 4 in. If the vertical scale is such that 1 in. repre¬ 
sents 40 lb. per sq. in., what is the mean effective pressure? 

(Ans., 30 lb. per sq. in.) 

Prob. 22. In a 10 by 20 in. double-acting engine at 100 r.p.m., 
with p m = 30, find the piston speed and ihp. 

(Ans., 333J ft. per min., 23.8 hp.) 

Prob. 23. A constant-speed 10 by 20 in. engine has a normal 
p m = 30 and a maximum pj = 60. What is its overload 
capacity? (Ans., 1.0 or 100 per cent.) 

Prob. 24. At 100 r.p.m., p m — 30. At 300 r.p.m., p m = 24. 
Find overload capacity if 100 r.p.m. is the normal speed. 

(Ans., 140 per cent.) 

Prob. 25. An engine rated at 100 hp. has an overload capacity 
of 210 per cent. What is the maximum hp. it can develop? 

(Ans., 310 hp.) 


12 


Thermodynamics, Abridged 


VALUES OF EXPONENTIALS IN CHAPTERS II AND III 



n 

1.3 

1.35 

1.4 

1.406 

1 

n 

0.769 

0.74 

0.714 

0.713 

71 — 1 

n 

0.231 

0.259 

0.286 

0.289 

0.355 0 - 231 = 0.787 


3.5 0 - 74 = 2.52 

0.9 0 - 74 

= 0.925 


40.259 = i.43 

(¥) 36 

= 6 


40.289 = 1.49 

2 7/6 

= 2.24 


4°- 74 = 2.78 

3 1.406 

= 4.68 


1Q-0.289 = o.514 

3.04 0 - 74 

= 2.28 


100-286 = 1.93 

3.37 0 - 259 

' = 1.37 


100.289 = 1.95 

3.37 0 - 74 

= 2.46 


100-714 = 5.18 

3.39 0 - 259 

= 1.38 


10 0 - 74 = 5.50 

3.39 0 - 74 

= 2.47 


11.08 0 - 269 = 1.86 

3.5 0 - 259 

= 1.38 


11.08 0 - 74 = 5.93 




CHAPTER II 


PERMANENT GASES. LAWS AND PROCESSES 
General Statement 

A clear understanding of the laws governing the behavior of 
permanent gases is absolutely essential to a proper comprehen¬ 
sion of steam machinery. Although those laws may seem for 
the next few pages to have little bearing on the heat-power 
equipment of a ship, their application will soon be shown. The 
few paragraphs which follow are as important in steam engine¬ 
ering as the multiplication table is in arithmetic. 

8. Gases. Thermodynamics deals, generally, with fluids 
rather than with solids, and (more narrowly) considers vapors 
and gases rather than liquids. A vapor is formed when a liquid 
evaporates. If the temperature of a vapor is greatly increased 
it becomes what we call a gas. Steam at 1200° is a gas. Air 
exists in the liquid form at about — 300° F. It might be con¬ 
sidered a vapor at — 250° F. In ordinary human experience, 
it is a gas. The present discussion deals with the “permanent” 
gases. Dry air, hydrogen, nitrogen, oxygen and a few of the 
compound gases are of this variety. Steam, NH 3 , C0 2 and S0 2 , 
on the other hand, are to be regarded as vapors. They will be 
examined later. In many thermodynamic calculations, great 
simplicity results (without any serious sacrifice of accuracy) 
from considering the permanent gases to be “perfect,” i.e., to 
conform exactly to certain physical laws which in reality are 
only approximated. 

The properties of gases which are chiefly to be considered are 
the pressure, volume and temperature. By pressure (p, lb. 
per sq. in.) we are to understand the uniform absolute (not 
gauge) pressure exerted per unit of surface by the gas on sur¬ 
rounding objects or by surrounding objects on the gas. The 

13 


14 


Thermodynamics, Abridged 


volume, v, is taken in cu. ft. The symbol t represents the uniform 
Fahrenheit temperature of the gas. 

9. Boyle’s Law. If the temperature of a gas is kept 

CONSTANT, THE PRESSURE VARIES INVERSELY AS THE VOLUME. 

— = — or V 1 V 1 — V 2 V 2 or vv — const, when t is const. (9) 
V2 

This is the equation of an equilateral hyperbola, referred to its 

asymptotes. When y= 14.696 
and t = 32°, v for 1 lb. of air 
is 12.387: hence at this tem¬ 
perature and for 1 lb. of air 
yx is always equal to 14.696 
X 12.387 = 182.04. Fig. 2 
represents Boyle’s law, the 
various curves being plotted 
for various values of the 
constant. 

Note that at sea level 
and normal barometer, y = 
GAUGE PRESSURE + 14.696. 
For ordinary calculations, 14.7 is near enough. 

Prob. 26. What is the volume of 5 lb. of air at 32° F. if the 
pressure is 10 atmospheres? (Ans., 6.1935 cu. ft.) 

Prob. 27. If the volume of 1 lb. of hydrogen is 100 cu. ft. 
when its pressure is 5.304 lb. by gauge, what will be the volume 
of 4 lb. of the same gas at the same temperature when its gauge 
pressure is 15.304 lb.? (Ans., 266f cu. ft.) 

Prob. 28. The gauge pressure on a gas tank is 85.304 lb., 
and its contents weigh 50 lb. What weight will it hold of the 
same gas at the same temperature under a gauge pressure of 
185.304 lb.? (Ans., 100 lb.) 

10. Charles’ Laws, (a) If the volume of a gas is kept 

CONSTANT, EQUAL INCREMENTS OF PRESSURE ACCOMPANY EQUAL 




















Permanent Gases — Laws and Processes 


15 


increments of temperature. Fig. 3 illustrates this. At 
constant volume, the relation between pressure and temperature 
is represented by a straight line. The equa¬ 
tion is, p = at + b, where a and b are con¬ 
stants. 

(b) If the pressure of a gas is kept 

CONSTANT, EQUAL INCREMENTS OF UOLVME 
ACCOMPANY EQUAL INCREMENTS OF TEMPER¬ 
ATURE. Again there is a straight line law: _ FlG * 3 ' Charle ® 

, . 7 TV & Law. (Constant Vol- 

v = ait + 6i: big. 4. 

& ume.) 



Prob. 29. The pressure in a closed tank 
is 200 lb. (absolute) at 32°, 400 lb. at 524°. What will it be at 
770°? (Ans., 500 lb.) 



Prob. 30. The absolute pressure of 
the gas in a balloon is 15 lb. at 40° 
and 30 lb. at 540°. If we disregard any 
increase in volume of the balloon, what 
is the pressure at 90°? (Ans., 16.5 lb.) 


Fig. 4. Charles’ Law. 11 • Absolute Zero. In Fig. 4, sup- 
(Constant Pressure.) pose the gas to be cooled indefinitely 

while conforming with the straight line 
law. At some very low temperature, the volume will then become 
zero. Experiments indicate a contraction of 1/492 of the original 
volume per degree of cooling if the gas is originally at 32° F. 
Hence zero volume would be reached after 492° of cooling, or 
at a temperature of — 460° F. This temperature (more accur¬ 
ately — 459.6° F.) is called the absolute zero. Experiments 
on the reduction of pressure at constant volume indicate the same 
point for the absolute zero. Hence Charles’ laws may be written, 


Vi 

V2 

Vi 

V2 


T i 


Ti 

T 2 


when v is constant 


when p is constant 


GO) 


Here T = t + 460 and denotes (as always hereafter) the 


ABSOLUTE TEMPERATURE. 








16 


Thermodynamics, Abridged 


No actual gas conforms exactly with the laws of Boyle and 
Charles. Zero volume would never be actually realized. The 
laws are close approximations for the behavior of “permanent’’ 
gases under normal conditions. 

Prob. 31. What is the location of the absolute zero on the 
Centigrade scale? (Ans., — 273° C.) 

Prob. 32. If an automobile tire did not expand, what would 
be the percentage increase in absolute pressure when the tempera¬ 
ture rose from 40° to 90° F.? (Ans., 10 per cent.) 

Prob. 33. The pressure in a closed tank is 200 lb. absolute 
when t = 32°. What is the pressure at 524°? Note Prob. 29. 

(Ans., 400 lb.) 

12. The Perfect Gas: Its Law. Conceive of a gas which con¬ 
formed rigorously with the laws of Boyle and Charles (Arts. 9, 
10, 11). Such a hypothetical substance is called the perfect 
gas. In many calculations, for dry air and its constituents and 
for hydrogen, these gases are treated as perfect, with a close 
approach to correctness. The behavior of the perfect gas may 
be expressed by a single equation 
which combines the laws already given. 

In Fig. 5, let the points 1 and 2 rep¬ 
resent any two states of a given weight 
of gas. Draw an equilateral hyper¬ 
bola 1-3 and a vertical line 3-2, inter¬ 
secting at some point 3. Along 1-3, 
t is constant and piVi = p & 3 . Along 
2-3, v is constant and p 2 /pz = T 2 /T z , or p 3 = p 2 {TalT 2 ). Then 
(since v z = v 2 ), PiVi = pzv 3 = p 2 V 2 {T z jT 2 ) or (since T z = Ti), 

PiOi V 2 V 2 pv 

rj i — ni or tft = constant. 

1 1 ^2 l 

If pv/ T = const., then pv = const, when T is const. — Boyle’s 
law. If piVi/Ti = P 2 V 2 IT 2 , then when p is const. vijv 2 = TjTo 
— Charles’ law. 

For air, when p = 14.696 and t = 32°, T = 492° and v per lb. 



Fig. 5 . Law of a Perfect 
Gas . 




Permanent Gases — Laws and Processes 


17 


is 12.387. Usual units employed are P = 144p, the pressure in 
lb. per sq. ft. Then Pv/T = (144 X 14.696 X 12.387) -f- 492 
= 53.36 or Pv = 53.36 T. For w lb. having a volume of V cu. ft., 
PV = 53.36 wT. For any gas, 

PV = wRT, (11) 

in which R is some constant. This is the equation of the 

PERFECT GAS. 

P = 144p = absolute pressure, lb. per sq. ft., 

V = volume, cu. ft., 
w = weight, lb., 

T = absolute temperature = t + 460. 

Equation (11) includes three variables. It is therefore the 
equation of a surface. A perspective representation of this 
surface appears in Fig. 5a. Any plane perpendicular to the OT 
axis cuts an equilateral hyperbola (like psxv) in the surface, 
the plane being a locus of constant temperature and the hyperbola 
representing Boyle’s law. Charles’ law for constant pressure is 
indicated by the straight lines map, Is, cut in the surface by 
planes perpendicular to the OP axis. If planes are drawn 
perpendicular to the OV axis, the intersections with the surface 
are straight lines xy, vben, which represent the other form of 
Charles’ law. 

Values of R for permanent gases are found as follows. Put a 
gas under “standard” conditions of pressure (14.696 lb. per sq. 
in. absolute) and temperature (32° F., or 491.6° absolute) and 
measure the volume of 1 lb. Call this volume Vq. Then 


PV 144 X 14.696r 0 
wT~ 1X491.6 


4.305^0- 


At any fixed condition, however, the volumes of gases are 

INVERSELY PROPORTIONAL TO THEIR MOLECULAR WEIGHTS, Or 

vo = c - 7 - m, where c is a number which has the same value for 
all gases, and m = molecular weight. The value of c may be 
determined from the volume and molecular weight of any gas. 



18 


Thermodynamics, Abridged 



Fig. 5a. Thermodynamic Surface for a Perfect Gas. 


Thus, for hydrogen (molecular weight = 2.016), Vo is known to be 
1 -7- 0.00562. Then c (based on standard conditions, as above 
defined) is 2.016 4- 0.00562 = 358.7. For any permanent gas, 
Vo = 358.7 -T- m. The value of R for any gas is then (4.305 
X 358.7) -r - m — 1544 -5- m. For hydrogen, R = 1544 4- 2.016 
= 766. The expanded form of the perfect gas equation is then 


pv _ 1544m r 
m 


( 12 ) 


The perfect gas is one having inert molecules: i.e., molecules 
which neither attract nor repel one another. Permanent gases 
approach this condition. 











Permanent Gases — Laws and Processes 


19 


Prob. 34. An automobile tire carries 100 lb. gage pressure 
at 40° F. and expands 5 per cent, when heated from 40° to 90° F. 
What is the pressure after heating? (Ans., 105.3 lb., gage.) 

Prob. 35. A balloon filled with hydrogen at 15 lb. per sq. in. 
absolute pressure has a volume of 766 000 cu. ft. at 80° F. 
What weight of gas does it hold? What weight of air at 14.696 
lb. pressure and 70° F. does it displace? When the volume 
increases 10 per cent, and the temperature of the gas is 90° 
what is the pressure? 

(Ans., 4000 lb.; 57 400 lb.; 13.9 lb. per sq. in. absolute.) 

Prob. 36. What is the temperature of air at 1000 atmospheres 
pressure (i.e., P = 1000 X 14.696 X 144) when it weighs 80.75 
lb. per cu. ft.? (Ans., 32° F.) 

Prob. 37. What is the ratio of sizes of tanks for an oxy- 
hydrogen light if both tanks have like pressures and tempera¬ 
tures? Note 2H 2 + 0 2 = 2H 2 0. Take molecular weight of 
oxygen at 32. (Ans., the oxygen tank is 50.4 per cent, the size 
of the hydrogen tank.) 

Prob. 38. A tank holds 560.4 lb. of nitrogen (molecular 
weight 28.02) at a pressure of 100 000 lb. per sq. ft. and a tempera¬ 
ture of 40° F. What is its volume? (Ans., 154.4 cu. ft.) 

Prob. 39. A 44 cu. ft. tank holds 10 lb. of gas at 794° F. 
and at a pressure of 10 000 lb. per sq. ft. What is the molecular 
weight of the gas? (Ans., 44.) 

Prob. 40. In the previous problem, after using gas from the 
tank the pressure is found to be 4400 lb. per sq. ft. and the 
temperature 140° F. What weight of gas has been removed? 

(Ans., 0.8 lb.) 

13. Mixtures of Gases: Dalton’s Law. Excluding chemical 
actions, the pressure of a mixture of gaseous bodies is the 

SUM OF THE PRESSURES AT WHICH THE VARIOUS CONSTITUENTS 
WOULD EXIST IF EACH IN TURN ALONE OCCUPIED THE SAME 
SPACE AS THE MIXTURE AT THE SAME TEMPERATURE. Let the 


20 


Thermodynamics, Abridged 


mixture condition be P, V, T, and let subscripts denote con¬ 
stituents. Then Dalton’s law is 

p p, P RiTwi R 2 Tw 2 

P - Pl + P* - ~y~ H f 

= y (RlWl + R 2 W 2 • • •) = — y- (13) 

where w = W\ + w 2 - • • and R refers to the whole mixture. The 
equivalent value of R in the equation PV — wRT is thus 

(RlWi + R 2 W 2 (Wi-p W2 • • •)• 

Prob. 41. What are the values of R for oxygen and nitrogen? 

(Ans., 48.25, 55.1.) 

Prob. 42. Considering air to be 3 parts oxygen to 10 of 
nitrogen, by weight, what should be the value of R for air? 

(Ans., 53.5.) 

Prob. 43. In the preceding, what part of standard atmos¬ 
pheric pressure (14.696 lb. per sq. in.) is due to the oxygen? 
(Ans., the pressure due to oxygen is 26.3 per cent, of that due to 
nitrogen and the oxygen pressure is 3.06 lb. per sq. in.) 

14. Specific Heats of Gases. A gas may absorb heat while 
remaining at constant temperature (Art. 20), the specific heat 
being then infinite; for specific heat is heat divided by change 
of temperature: or it may change in temperature without 
absorbing heat (Art. 28), when the specific heat is zero. That 
this must sometimes be the case is evident from the fundamental 
equation, H = T + I + W. If H is + and (W + I) is + and 
greater than H, T must be negative. We cannot assign numeri¬ 
cal values for the specific heats of gases unless at the same time 
we stipulate the type of process with which such specific heats 
are realized. 

Two standard processes are clearly recognized. The gas may 
be kept at constant pressure. The corresponding specific 
heat, called k (value 0.2375 for air), will then be fairly constant 
over ordinary temperatures. Second, the gas may be kept at 





Permanent Gases — Laws and Processes 


21 


constant volume. In this case, the (reasonably) constant 
specific heat is called l, the value for air being 0.1689. At very 
high temperatures, k and l are rapidly variable. 

15. Difference of Specific Heats. When a gas is kept at con¬ 
stant volume, it can do no mechanical work, for there is no 
such work possible without mass-motion. If heated at constant 
volume, energy will be stored up: so will it be stored when the 
gas is raised in temperature at constant pressure. If at constant 
pressure mechanical work is done, then the difference between 
k and l is due wholly to the performance of work in the one case 
and not in the other, and k will be greater than l. 

16. Work Done at Constant Pressure. Mechanical work is 
done when a force acts over a distance. If the force is in lb. 
and the distance in ft., the work, in ft. lb., is force X distance. 
Take the case of a gas, at the pressure Pi, having a free (movable) 
surface of A sq. ft. Let expansion occur in a direction perpen¬ 
dicular to such surface, and let the linear movement in this 
direction be L ft. The volume of the gas must increase during 
the movement, by an amount equal to the space swept through 
by the moving surface. This space is AL cu. ft. If F 2 and V\ 
are final and initial volumes in cu. ft., then AL = F 2 — Vj and 
W = PiAL = Pi(F 2 — Fi) ft. lb. Thus the external me¬ 
chanical WORK, IN FT. LB., DONE BY A GAS KEPT AT CONSTANT 
PRESSURE IS EQUAL TO THE PRESSURE (LB. PER SQ. FT.) MULTI¬ 
PLIED BY THE INCREASE IN VOLUME (CU. FT.). 

Suppose the initial and final temperatures to have been T 1 
and P 2 . Then by Charles’ law 


F 2 _ T2 V _ 2 
Fi “TV Fi 


=r- 1 = 



T 2 V 2 -V l _T2-T l 

rn L t 7 rn 


and 



w = P l (V 2 - Fi) = ~ (r ‘ > - - Tl) - 


Hence 





22 


Thermodynamics, Abridged 


But PiFi = wRTi, hence 


PiPT 

Then W = wR(T 2 - Pi). Then 

ff = T+J+^= T+J + 


wR(T 2 — Pi) 
778 


which we may put equal to kw(T 2 — Pi) since the process is 
one of constant pressure. Note that P 2 and Pi are tempera¬ 
tures, while T is the heat consumed in changing temperature, 
as in Art. 2. 

For heating at constant volume, on the other hand, W = 0 
and H — T + I = lw(T 2 — Pi). This expresses the value of 
T+ I for any process from a state i to a state 2 . The value of 
P+I depends on the two states only, and not on what happened 
in moving from one to the other of them. See Art. 18. Apply¬ 
ing this value of T + J to a process at constant pressure, 

kw(T 2 - TO = MT 2 - TO + ~ (r, - TO 

or 

7S- < 14 > 


Thus the quantity R in the perfect gas equation is simply 

THE DIFFERENCE OF THE TWO SPECIFIC HEATS expressed in ft. lb. 

It is the external work done in heating 1 lb. of gas 1° at constant 
pressure. 

Prob. 44. Given k for hydrogen = 3.409, find l. 

(Ans., 2.424 B.t.u.) 

Prob. 45. What value of R is indicated for air, with k and l 
as given? (Ans., 53.37.) 

Prob. 46. 1 lb. of air at 69° F. is placed in a vertical cylinder. 

A piston without weight rests above this air. There is no 
friction, so that the pressure on the air is that of the standard 
atmosphere above, or 14.696 X 144 = 2116.2 lb. per sq. ft. 




Permanent Gases — Laws and Processes 


23 


The cylinder cross-sectional area is 1 sq. ft. State the pressure, 
volume and temperature of the air in the cylinder, using standard 
units. (Ans., P = 2116.2, V = 13.34, T = 529.) What will 
be the volume after heating this air 1 °? The mechanical work 
done during such heating? (Ans., 13.365 cu. ft.: 53 ft. lb.) 

17. Ratio of Specific Heats. The ratio of k to l , called y, is 
of very great importance in Thermodynamics. Its value for 
air is 0.2375 -5- 0.1689 = 1.406. For all of the permanent gases, 
its value is close to 1.4. The following relations should be noted: 

1 j _ R k _ R 

778’ l ~ y ’ 1 ~ 778(2/ - 1) ’ 

7 R y 778ifc 

k ~ 778(y-l)’ y ~ 778k - R’ iJ - 778 %~ 1 )- 

Prob. 47. Find y for hydrogen. (Ans. 1.406.) 

Prob. 48. If for nitrogen k = 0.2438, make two estimates of 
the probable value of 1. (Ans., taking R = 55.1, Z = 0.173. 
Taking y = 1.4, l = 0.174.) 

18. Joule’s Law. It has been demonstrated experimentally 
that with perfect gases the amount of disgregation work is so 
small that it may be ignored. For the permanent gases, 

UNDER NORMAL CONDITIONS AND WITH SUFFICIENT ACCURACY 
FOR MOST ENGINEERING APPLICATIONS, 7=0 and H = T + W. 
This refers to Equation (3). In a perfect gas, I would be 
absolutely zero (Joule’s law). 

It follows from Joule’s law that H = T during a constant 
volume (zero work) operation. A constant volume process may 
therefore be employed to measure the T effect per unit of heat 
supplied. It is found that H is l B.t.u. per lb. per degree of 
temperature change in such a process. Then for any process 
whatever, H = lw(T 2 — 7\) + W, where W is expressed in 
B.t.u. This principle underlies much of what follows. For a 
given weight of a given gas, the W effect depends on the process 
and may range from zero upward: but the T effect is determined 
solely by the temperature change, no matter what the process is 
or how the change is effected. 

3 





24 


Thermodynamics, Abridged 


Prob. 49. The temperature of 8 lb. of air is raised 125° while 
mechanical work equivalent to 100 B.t.u. is done by the air. 
How much heat is supplied? (Ans., 268.9 B.t.u.) 

Prob. 50. By what type of process can the temperature of a 
gas be increased at the least expenditure of heat? (Constant 
' volume process.) What is this expenditure for 100 lb. of hydro¬ 
gen heated 10°? (Ans., 2424 B.t.u.) 

Prob. 51. In an air compressor, 778 000 ft. lb. of work are 
done on the air, which meanwhile rises 1000° in temperature. 
If 5 lb. of air are present, state values of T, I, W and H. 

(Ans., T = + 844.5, I = 0, W = - 1000, H = - 155.5.) 


Thermodynamic Properties of Permanent Gases 


Gas 

Symbol 

Boiling 
Point at 
Atmospheric 
Pressure, 
Deg. Fahr. 

Molecular 

Weight 

R 

k 

l 

— 

y 

Helium. 

He 

-450 

4.0 

386 

1.25 

0.75 

1.667 

Hydrogen. 

Ho 

-412 

2.016 

766 

3.409 

2.424 

1.406 

Nitrogen. 

n 2 

-318 

28.02 

55.1 

0.2438 

0.173 

1.409 

Oxygen . 

0 2 

-297 

32 

48.25 

0.2175 

0.1555 

1.399 

Air. 


-314 

— 

53.36 

0.2375 

0.1689 

1.406 

Methane. 

ch 4 

-265 

16.032 

96.31 

0.5929 

0.4698 

1.26 

Carbon monoxide. 

CO 

-313 

28 

55.14 

0.2425 

0.1716 

1.413 


AT , D 1544 , . . R 

Note: R =- and l = k — . 

m 778 

19. Graphical Representation of External Work. It has been 

shown that the external work done at constant pressure is equal 
to the pressure multiplied by the increase of volume. In Fig. 6, 
let ab represent a constant-pressure process. Horizontal dist¬ 
ances represent volumes. Vertical distances represent pressures. 
Then W = PabiYb — V a ) = the area abdc. If the volume 
decreases, as when the process is from b to a, the work is 
represented by the same area, but it is work done ON the gas, 
and has a negative sign. 

In Fig. 6, consider any path, ef. Take a short portion, gh , 
of this path. When gh is infinitely short, it may be considered 
horizontal. Then W gh = ghkj. The work done along the whole 




















Permanent Gases — Laws and Processes 


25 


path ef is the sum of such areas as ghkj, or is the area efml. If 
the path is from left to right (e to/), W is +; otherwise, it is —. 
The following theorem may be stated: 



On a PV diagram, the value of W for any process is 

REPRESENTED BY THE AREA BELOW THE LINE REPRESENTING 
THAT PROCESS. 

For a constant-volume process, no, Fig. 6, there is no sub¬ 
tended area and hence W = 0, as already inferred. 

20. Isothermal Process. An isothermal operation is one in 
which THE TEMPERATURE REMAINS CONSTANT. The gas must 
then conform to Boyle’s law and the process is represented (Fig. 6) 
by an equilateral hyperbola pq. The area below this curve, 
yqrs, represents W . The magnitude of this area can be obtained 
by integration, the equation of the curve being P p V p = P q V q . 
It is 

w= P p v„ log. ^ ft. lb., 

V P 


( 15 ) 






26 Thermodynamics, Abridged 

where log e = 2.303 logio. The symbol log (without a subscript) 
means logio* 

In Fig. 5a, psxv, ab, mlyn, are isothermals. 

For an isothermal process, the fundamental equation, II = T 
+ I + IF, becomes H = W: for T = 0 for an isothermal and 

I = 0 for a permanent gas. To obtain II in B.t.u., divide W 
by 778. 

Prob. 52. If 10 lb. of air at standard atmospheric pressure 
and 32° F. are heated at constant pressure to 524° F., find 
H and IF. (Ans., H = 1168.5 B.t.u., W = 262 134 ft. lb.) 

Prob. 53. In the previous problem, show that numerically 

II - (IF/778) = lw{T 2 - 2P0. 

Prob. 54. If 10 lb. of air at standard atmospheric pressure 
and 32° F. are expanded isothermally to twice the original 
volume, find IF and H. (Ans., IF = 181 730 ft. lb.: II = 233.6 
B.t.u.) 

Prob. 55. If 10 lb. of air are cooled at constant volume from 
524° to 32°, find IF and II. (Ans., IF = 0, II = — 830.99 B.t.u.) 

21. Cycle. Most thermodynamic operations consist of a series 
of processes which restore the fluid to its original condition. 
Such a series constitutes a cycle, which is represented by a 
closed figure on a PV diagram. For an entire cycle, the funda¬ 
mental equation H — T + I + IF becomes H = W: because 
T — 0 when the original temperature is regained and I = 0 for a 
permanent gas. Hence, 

In a closed cycle, the net absorption of heat 

EQUALS THE NET EXTERNAL WORK DONE. 

217 and 2 IF must be expressed in the same units, either heat 
units or foot-pounds. This refers to a clockwise cycle. If the 
cycle is counter-clockwise, the law holds, but ZH and 2 IF are 
both negative. The area of the cycle itself is the algebraic sum 
of subtended areas under the paths which compose it: hence the 
area of the cycle represents 2 H and 2 IF. 

The area of a closed cycle on a PV diagram 

REPRESENTS THE NET AMOUNT OF HEAT ABSORBED OR 


Permanent Gases — Laws and Processes 


27 


EMITTED AND ALSO THE NET AMOUNT OF EXTERNAL 
WORK DONE BY OR ON THE SUBSTANCE: both in ft. lb. 

Prob. 56. In the cycle abc of Fig. 7, for 10 lb. of air, find the 
external work done along each path and the heat absorbed along 
each path and show that 271 = 2TF. See Probs. 52-55. (Ans., 
H ab = 1168.5, H bc = - 830.99, H ca = - 233.6, 217 = 103.9 
B.t.u. = 80 400 ft. lb.: W ab = 262 134, W bc = 0, W ca = - 
181 730, 2IF = 80 400 ft. lb.) 

22. Efficiency of Cycle. The object of a thermodynamic oper¬ 
ation is the performance of some net amount of external work. 
This is the result of the reception of heat. Since efficiency 


= EFFECT -f* CAUSE, 

The efficiency of a cycle is the net amount of 

HEAT ABSORBED OR EXTERNAL WORK DONE (AREA IN 

B.t.u.), divided by the whole amount of heat 

RECEIVED. 

In other words, it is 277 divided by the sum of all of the 77 terms 
having + signs. In Fig. 7, it is abc in B.t.u. divided by Hah- 


Prob. 57. Find the effici¬ 
ency in Fig. 7. 

(Ans., 104 -T- 1168.5 = 0.089.) 

23. Clearance, Displace¬ 
ment. If such a cycle as 
that of Fig. 7 were actually 
employed, the displacement 
(per 10 LB. OF AIR USED) 
would be V c — V a — 123.87 
cu. ft. The clearance (Art. 
7) would be V a -r- (V c — V a ) 
= 1.0, or 100 per cent. The 

MEAN EFFECTIVE PRESSURE 

(Art. 7) would be p m = abc 



123.87 = 653 lb. per sq. ft. or 


- (Vc - Va) = (104 X 778) ^ 

4.54 lb. per sq. in. If 100 lb. of air were used per min., the engine 
would develop (100/10) X (80 400/33 000) = 24.5 hp. (see Prob. 
56). 










28 


Thermodynamics, Abridged 


Prob. 58. In the cycle of Fig. 8, find clearance, p m , displace¬ 
ment and hp. if 50 lb. of air are used per min. and the cycle 
represents 1 lb. of air. State the abso¬ 
lute temperature at each point. (Ans., 
clearance = 0.333: displacement =26.68 
cu. ft. per lb. of air or 1334 cu. ft. per 
min., p m = 20 v 83 lb. per sq. in., hp. = 
121.273, Td = 500°, T a = 1000°, T b = 
4000°, T c = 2000°.) 

24. General Case of Gas Process: the 
Poly tropic. Practically all heat engine 
processes may be represented on the PV 
diagram by smooth curves, V increas¬ 
ing as P decreases. Fig. 9 shows several such curves. An 
equation which represents them is PV n = const, (the exponent 
applying to V only). When several such curves (called poly¬ 
tropic curves) pass through a given point, like a, the value of 
n indicates the steepness of the curve. Equality of n values 
does not imply coincident curves. For n — 1, for example, one 
curve is plotted through a and another through b. 

For n = 0, PV° = const., or P = const. — the process is one 

of CONSTANT PRESSURE. 

For n = 1, PV = const. — the process is isothermal. 

For = qo, PV 00 = oo, P 1/oo F — const., V = const. — the 
operation is at constant volume. 

25. Relations of Properties in a Polytropic Process. Consider 
a case of polytropic ( PV n = const.) expansion from 1 to 2. 
Then P\VV = P 2 F 2 n . The substance being a gas and the 
weight constant, the perfect gas equation may be written P\V\T 2 
= P 2 V 2 T\. Dividing the second of these equations by the first, 

Y" n 

Ti \Vj • (16) 

Extracting the n root of the first equation, and then dividing 
the second, by it, and transposing, 


P 

6000 

3000 


d 

i 




Fig. 8. Problem. 











Permanent Gases — Laws and Processes 


29 


When n = 0, Equation (16) reduces to Charles’ law for con¬ 
stant pressure. When n = co, Equation (17) reduces to Charles’ 
law for constant volume. 



Fig. 9. Polytropic Processes. 


Thus suppose T\ = 492, Pi = 2116.2, V\ — 12.387. Let 
n = 1.4, V, = 123.87. Then T 2 = 492 X KT 04 , log T, = 2.692 
_ o.4 = 2.292, T 2 = 195.9. Also (since P 2 /Pi = (Vi/V 2 ) n ), 
P 2 = 2116.2 X (1/10) 1 - 4 , log P 2 = 3.3256 - 1.4 = 1.9256, P 2 





















30 


Thermodynamics, Abridged 


= 84.26. As a check, from P\V\T 2 = P*V 2 Ti, T 2 = (84.26 
X 123.87 X 492) (12.387 X 2116.2) = 195.9. 


Prob. 59. Given P x = 10 000, V x = 10, V 2 = 50, n = 2. 
The substance is 4 lb. of nitrogen. Find T i, T 2 , P 2 . 

(Ans., Ti = 453.72, T 2 = 90.74, P 2 = 400.) 


26. External Work, Polytropic Process. The area under a 
curve passing through a high-pressure point a and a low-pressure 
point b, if the equation of the curve is P a V a n = PbVb n , is by 
integration, 


_ PgVg - PbVb 

71—1 


( 18 ) 


This gives the work in ft. lb., which is to be regarded as + if the 
process is from a to b (left to right). If the reverse, IF is —. 
For n = 1, the work is indeterminate by this method, and 
Equation (15) should be used. 

Prob. 60. In the problem immediately preceding, how much 
external work is done? (Ans., IF = + 80 000 ft. lb.) 


27. Heat Absorbed, Polytropic. In any gas process, H = T 
+ IF. For all processes, T = lw(T 2 — T i). For a poly tropic, 


IF = 


71—1 


From Equation (11), P X Vi = wRTi and P 2 V 2 = ivRT 2 . Hence 

Hence for a polytropic, 

Noting that 

and putting k = ly, 

H = Ti). (19) 







Permanent Gases —Laws and Processes 


31 


But for any process, if the specific heat is s, H = sw(T 2 — T i). 
Hence for a polytropic, s = l(n — y)/(n — 1). The value of s 
for a given gas depends on n alone. The polytropic is then a 
curve of constant specific heat. This is why the polytropic 
is so important. Practically all permanent gas processes are 
represented by poly tropic curves. So also are many vapor 
processes. Many practical applications will be found in Chapter 
III. For n = 0, s = ly = k, as is to be expected. If n > 1 < y, 
s is negative, i.e., heat is emitted when the temperature 
rises or absorbed when the temperature falls. This actually 
occurs in every air compressor. During compression, the 
temperature of the air rises although the jacket water is con¬ 
tinually abstracting heat from the air. The reason is that we 
are doing external work on the air faster than we are absorbing 
heat from it. For positive values of s, a rise of temperature 
indicates reception of heat. When n = 1, s = oo; a condition 
necessary for an isothermal process, in which no finite supply 
of heat can ever change the temperature. 

Prob. 61. In Prob. 59, what is the specific heat? (Ans., 
0.0998.) How much heat is absorbed or emitted? (Ans., 
H = - 145 B.t.u.) What is the value of T in H = T + / + Wt 
(Ans., - 247.8 B.t.u.) 

28. Adiabatic. An adiabatic process is one during which no 
heat is received or emitted by the fluid in question. Under 
adiabatic conditions, H = T + / + W = 0: but T and W are 
both finite. (If the substance is a permanent gas, / = 0.) If we 
assume that the adiabatic may be represented by an equation 
in the form PV n = const., n being undetermined as yet, 


W ~ — T or 


P iFi - P 2 V 2 _ lw(Ti - T 2 ) 
n- 1 778 9 


where the process is from state 1 to state 2 . Since PiV 1 = wRT 1 
P 2 V 2 = wRT 2 , t his leads to ' = 778 R/(n — 1). In Art. 17, it 
was shown that l = 778 R/(y - 1). Hence for an adiabatic 
process n = y. The equation of the adiabatic is P\V\ v = P 2 V 2 V 
and for such process, following Arts. 25, 26, 




32 


Thermodynamics, Abridged 


t 2 _ (V2\- y _ /p 2 y y - i)/y 

VpJ 

yy, P\V l ~ P 2 V 2 

kK =- 1 -. 

y - i 

During an adiabatic expansion, the temperature falls. In Fig. 
5a, ab is an isothermal, ae an adiabatic. The two curves pro¬ 
jected to the PV plane appear as zk, zg. 

Prob. 62. Air expands adiabatically from Pi = 10 000, V\ 
= 10 to V 2 = 30. Find P 2 and W i2 . 

(Ans., P 2 = 2140, W 12 = 88 100 ft. lb.) 

29. Summary. The following table summarizes what has 
been learned concerning permanent gas processes. The values 
of H are in B.t.u.: those of W are in ft. lb. if P is taken in lb. 


Permanent Gases—Processes 


Name 

Constant 

Property 

H 

w 

5 

n 


Horizontal . . . 

Pressure 

kw{T 2 - Ti) 

Pn(V 2 - 70 

k 

0 


Vertical. 

Volume 

1w(T 2 - TO 
PiV 1 lOge ^ 

0 

l 

00 


Isothermal. . . 

Temperature 

Vo 

P t V! lOge A 2 
Vi 

1 


778 

00 


Adiabatic. . . . 

Entropy* 

0 

PiVx - P 2 V 2 

y - 1 

0 

y 

t 2 

Ti 

Poly tropic.. . . 

Specific heat 


Pi7: - P 2 V 0 

n— 1 


t 2 


n - 1 


T\ 


Remarks 


VtlVi = T 2 /T 1 
P 2 /P 1 = T a /T x 


P 1 V 1 = P 2 V 2 



* The meaning of this term may be disregarded for the present. 


per sq. ft. Only the last line of the table need be emphasized. 
With the exceptions encircled, all of the necessary relations may 
be derived from those of the polytropic by substituting the 
proper value for n. 

30. Application. A cycle will now be worked out in detail to 
illustrate these formulas and to show how a complete check of 


















Permanent Gases —Laws and Processes 33 


the computations is possible. In Fig. 10, data are as follows: 
Clockwise cycle, 2 lb. of nitrogen 


Pi = P‘i= '140’000, 

/w = p 3 iy, 

Vi= V 5 = 0.5, 

T 2 = 1270.32, 

P 5 = 80 000, 

PoFo = P 4 F 4 , 

P 4 = 20 000, V 3 = 3, 

P4l r 4 n = P 3 P 3 ”, with n 

unknown. 

The unknown properties 
are first found: 


—-/- 


\fanl 

0 

1 

Presi 



S' 

L 










_r 

1 

V 

\c 

\ 

L. 

Is- 



O 



%• 

\o 












/ 







£ 



Fig. 10. Cycle for Computation. 


P 1 F 1 140 000 X 0.5 

wR ~ 2 X 55.1 


635.16, 


T b = T l 


Ps 

Pi 


= 635 X 


80 000 
140 000 


363 


T 


4, 


T 1270 

V '-V'Tr °- 5 X 635 = 1.0, 



80000 = 

U ' 5 X 20 000 / ' U ’ 


/J/ \ 1.409 / 1 \ 1-409 

p 3 = p 2 ( F -;j = 140000 ( 3 ) , 

/ F, \-°- 409 

= r Jt l pr ) = 1270 X 3-°- 409 , 

log P 3 =[5.1461 - 0.6723 = 4.4738, P 3 = 29 770, 
log T, = 3.1040 - 0.1951 = 2.9089, T 3 = 810.5. 



























34 Thermodynamics, Abridged 

Also, to check, 


T* = 


P 3 F 3 29 770 X 3 


= 810.5. 


wR 2 X 55.1 
To find the value of n for the process 4 — 3: 

P 4 F 4 n = P 3 F 3 n , log P 4 + n log V 4 = log P 3 + n log F 3 , 

log P 3 — log P 4 
* - log V 4 - log Vs’ 

4.4738 - 4.3010 

n ~ 0.3010 - 0.4771 “ 0-98L 

The negative sign is unusual: note the slope of the curve. 
It does not affect the method of procedure. The specific heat 
for the path 4 — 3 is 



S 4 3 = l 

n — 

n — 

y - 0 173 X ~ 2,390 - + 
i U * 17 " X - 1.981 + 

0.2087. 


We 

now compute 

the heat quantities: 



Hn = 

wk(T 2 — 

TO 

= 2 X 0.2438(1270 - 635) 

= 309.7 

(+) 

ff 23 = 




0 


H 3 4 = 

ws(T 4 — 

TO 

= 2 X 0.2087(363 - 810.5) 

= 186.8 

(-) 

7T _ 

P 5 F 5 log, 

Vi 

” v v 

80 000 X 0.5 X 2.3 log 4 

_ 719 

( \ 

J * 4 5 — 

778 


778 

— / l.Z 


Hs 1 = 

wl(Ti - 

TO- 

= 2 X 0.173(635 - 363) 

= 94.1 

(+) 





= 146 

B.t.u. 


The external work quantities are: 

w 12 = P 12 (F 2 - Vi) 

= 140 000(1.0 - 0.5) 
P2V2 - p 3 f 3 


70 000 (+) 


W‘ 


y - 1 

(140 000 X 1.0) - (29 770 X 3) 
0.409 


= 123 900 (+) 













Permanent Gases — Laws and Processes 


35 


w 34 = 


P3F3 - P4F4 

n — 1 


- (29770 X -L981 0000 X 2 ) = 24900 (-) 
^45 = PJ 7 5 log, 

= 80 000 X 0.5 X 2.3 log 4 = 55 400 (-) 
W si = 0 


2 JF = 113 600 ft. lb. 

= 146 B.t.u. 


Thus 2 IF = 277. 

31. Isodiabatics are polytropic curves not coincident, but 
having the same value of n. In Fig. 9, the lines M and N are 
isodiabatics. The following theorems will be found later to 
save much time in computation with many common forms of 
cycle: 

1 . Let a pair of isodiabatics, M and N, Fig. 11, be cut by lines 
of constant pressure, Pi and P 2 . Then 


P e V e n = P g V g n ; P f V f n = P h V h \ 


Dividing the first of these equations by the second, 


p v n P V n 

py? = pltv or since P e = P f and P a = P h , 


Fe _ — = — = h 

V, V h Tf TV 


( 20 ) 


the equality of temperature ratios arising from Charles’ law. 

2 . Let a pair of isodiabatics, M and N, be cut by lines of 
constant volume, Fi and F 2 . Then 


PjVj n = PiV : n , P k V k n = P m V m n , 

PjVp _ PiVp . Pj _ Pi _ Tj _ Tj_ 
P k v k n P m V m » 0r Pk P,n T k TJ 


Charles’ law being again applied. 


( 21 ) 












36 


Thermodynamics, Abridged 


P 



Fig. 11. Isodiabatics. 


3. Let a pair of isodiabatics, M and N, be cut by lines of 
constant temperature, Ti and T 2 . Then 

Ta _ (Pa \ (n - 1),n _Tc_( P 5 \ (n - 1) ' n 
T b \PJ T d \pj 

Pa P. Pa = Pt = V, 

P b Pd’ Pc Pd V a V,,’ 

Boyle’s law being applied in this instance. 


( 22 ) 










Permanent Gases — Laws and Processes 


37 


Nearly all heat engines work in cycles made up of one or more 
pairs of isodiabatics (see, for example, Art. 38). 

Prob. 63 . In Fig. 11, given t k = 40° F., t 5 = 1040° F., 
P k = 20, Pi = 27.3, find Pj and P m . (Ans., Pj = 60, P m = 9.1.) 

Prob. 64 . In Fig. 11, if V e = 15, V f = 30, t f = 904° F., 
v h = 36, tg = 260° F., find V g , T h and T e . 

(Ans, V g = 18, T h = 1440, T e = 682.) 

Prob. 65 . In Fig. 11, P d = 10, P b = 26.5, V d = 74, V c = 69, 
T c = 750, T d = 575. Find V b , P c , P a , V a . 

(Ans, V b = 27.9, P c = 14, P a = 37.1, F a = 26.) 

32. Imperfect Gases: Application to a Gun. An imperfect 
gas may be conceived as a perfect gas through which there are 
distributed incompressible particles. Let the volume of 1 lb. 
of such particles be a cu. ft. Then the remainder of the gas 
conforms to Equation (11), or 

P(V - wo) = wRT. (23) 

This is the Van der Waals equation for imperfect gases. The 
gas liberated in the powder chamber of a gun furnishes a good 
example of an imperfect gas. For powder gases, the value of a 
is approximately 0.00H 0 , where v 0 = specific volume at 32° F. 
and normal barometer. Hence 


a = 0.00 Ho = 


492 R 

14.696 X 144 


0.000232E. 


Prob. 66. Find a when R = 46. (Ans, 0.0107.) 


Prob. 67. The powder chamber of a gun has a volume of 
8.5 cu. ft. The weight of powder is 300 lb, the temperature 
1000° F. before the projectile moves. Assuming R for powder 
gases = 46, find the powder pressure at the moment the pro¬ 
jectile begins to move. Compute first by Equation (11) and 
afterward by Equation (23). (Ans, p = 16 400 if the gas is 
perfect: p = 26 500 by the Van der Waals equation.) 



38 


Thermodynamics, Abridged 


Prob. 68. If the gun in the foregoing problems is 12-in., 
50 ft. long from seat to muzzle, what is the whole volume of gas 
behind the projectile when it emerges from the muzzle? 

(Ans., 47.77 cu. ft.) 

Prob. 69. If maximum pressure occurs when the projectile 
has traveled half way from seat to muzzle, what is the volume 
of gases behind the projectile then? (Ans., 28.13 cu. ft.) If 
maximum pressure occurs when the projectile has moved 19.6 ft., 
what is the volume at that point? (Ans. 23.88 cu. ft.) 

Prob. 70. In the second case of the foregoing problem, 
assume that a plot of pressure against volume would be a straight 
line from atmospheric pressure and 8.5 cu. ft. volume to 30 000 
lb. per sq. in. pressure at 23.88 cu. ft. volume. How much 
external work is done up to the point of maximum pressure? 

(Ans., 33 200 000 ft. lb.) 

Prob. 71. In the foregoing, the further expansion of the 
powder gases is polytropic, with n = 7/6. What is the pressure 
at the muzzle? (Ans., 13 400 lb. per sq. in.) How much 
external work is done from point of maximum pressure to 
muzzle? (Ans., 66 000 000 ft. lb.) 

Prob. 72. In all of the foregoing (Probs. 70, 71) if the powder 
contains 1620 B.t.u. per lb., what is the thermodynamic efficiency 
of the gun? (Ans., 0.26.) 

Prob. 73. Ignoring frictional and other losses, the total 
mechanical work becomes ^ MU 2 where M = mass of projectile 
(= lbs. weight 4- 32.17) and U = muzzle velocity, ft. per sec. 
Find muzzle velocity if the projectile weighs 800 lb. 

(Ans., 2820 ft.) 

Prob. 74. In the preceding problems, how long would the 
gun need to be in order to reduce the muzzle pressure to that of 
the atmosphere? (Ans., about 4 miles.) 

Prob. 75. Under the conditions adopted above, find the 
temperature (by the Van der Waals equation) at the point of 
maximum pressure and at the muzzle. (Ans., 6500°: 6200°: abs.) 


Permanent Gases — Laws and Processes 


39 



33. Graphical Representation of Heat Absorbed. In Fig. 12, 
let ab represent any permanent gas process. Draw two adi- 
abatics through a and b. At „ 
an indefinite distance to the 
right the two adiabatics will 
meet. Designate the meeting- 
point as M. Then the figure 
abM is a cycle and the area 
abM, as already shown, rep¬ 
resents the net amount of 
heat absorbed or rejected in 
that cycle. There are only 
three processes included in 
the cycle: Ma, ab and bM. 

We know by definition of an 
adiabatic that H Ma — 0, II hM = 0. 
sented by the area abM : or 

The heat absorbed or emitted along any path 

IS REPRESENTED BY THE AREA INCLUDED BETWEEN 
THAT PATH AND TWO ADIABATICS PASSING THROUGH ITS 
EXTREMITIES AND PROLONGED INDEFINITELY TO THE 
RIGHT. 

If, as viewed from a point lying between the adiabatics, the 
path is from left to right, H is +. Otherwise, H is —. Along 
an adiabatic, by this theorem, H = 0. 


Fig. 12. Heat Absorbed, Any Process. 
Hence H a b alone is repre 


34. Carnot Cycle. In Fig. 13, a cycle abdc is formed by the 
combination of a pair of isothermals, ab and cd, with a pair of 
adiabatics, ac and bd. As in all cycles, the efficiency is net 
heat -f- gross heat. The net heat is the area abdc. The 
“gross” or plus heat is made up of four possible terms: but of 
these H ac = Hbd = 0, the paths in question being adiabatic: 
while H dc is eliminated because for a clockwise cycle its value is 
negative. Hence the efficiency is 

abdc Hab Hdc 

6 mabM H a b 


4 






40 


Thermodynamics, Abridged 


PaV a l0g e P ~ P C V C log, ~ 

la ' c 


PaV a foge 


n 

V a 


From Equation (22), Vb/V a = Vd/V c : hence 

PaV a - P C V C T a ~ T c T, ~ T 2 


P V 

x a r a 


T 

1 a 




(24) 



where T\ and T 2 are the highest and lowest (absolute) tempera¬ 
tures of the cycle. 

This value is realized in a Carnot cycle, no matter 

WHAT GAS IS USED. 

Between assigned limiting temperatures, no possible 

CYCLIC PROCESS CAN HAVE AN EFFICIENCY EXCEEDING THAT OF 

the Carnot cycle. In Fig. 13, if the proposed superior cycle 









Permanent Gases — Laws and Processes 41 

were abde, the point e would be below the lower limiting tempera¬ 
ture. If it were afdc, the point / would lie above the upper 
limiting temperature. Neither condition is allowable under the 
stipulation made. An allowable modified cycle would be abgc. 
As compared with the Carnot, this gives a reduced amount of 
external work represented by. the area cgd : but the gross heat 
absorbed is still mabM and the efficiency is hence reduced. 
Consider next the cycle hbdc. It gives less work than the 
Carnot, represented by the area abh : but it s^ves heat, the 
gross heat being now mhbM. The saving of heat is exactly 
equal to the loss of work. Then the efficiency of the new cycle is 

abdc — abh 
mabM — abh' 

Since abdc/mabM < 1.0, the efficiency of the new cycle is less 
than that of the Carnot cycle. 

If some such form as abjdc were proposed, it would be sur¬ 
passed in efficiency by a broadened-out Carnot cycle aklc, and 
hence by the original cycle abdc, which lies between the same 
temperature limits as aklc. 

No HEAT ENGINE CAN HAVE AN EFFICIENCY EXCEEDING 
( T i — T 2 ) / T i WHERE T i AND T 2 ARE THE HIGHEST AND LOWEST 
ABSOLUTE TEMPERATURES OF THE FLUID IN THE ENGINE. 

Equally definite, but lower, limits of efficiency will presently 
be established for various specific types of engine. 

Prob. 76. x4 Carnot cycle gives, area = 100, gross heat = 200. 
A competitive cycle hbdc is formed, with abh = 20. State the 
efficiency of each cycle. (Ans., 0.50 and 0.444.) 

Prob. 77. What is the least amount of coal (12 725 B.t.u. 
per lb.) per hour per horse power used for an engine using steam 
at 340° F. and exhausting to a condenser at 140° F.? 

(Ans., 0.8 lb.) 

Prob. 78. The maximum temperature in an oil engine is 
3500° abs., the exhaust temperature 1000° abs. The engine 
develops 190 hp. and uses oil of 20 000 B.t.u. per lb. What is 
the least possible oil consumption per hr.? (Ans., 33.8 lb.) 


42 


Thermodynamics, Abridged 


35. Carnot Cycle Reversed. In Fig. 13, suppose the cycle to 
be worked counter-clockwise. Begin at c. Suppose the gas to 
be very cold—say at 0° F. Let it pass through a room, receiving 
heat (area mcdM) but not increasing in temperature. The 
point d is reached. Then compress the gas adiabatically to b. 
Its temperature will rise—say to 150° F. Then pass it across 
coils through which cold water is circulating. Heat will be 
abstracted (area mabM ) but the temperature may be kept con¬ 
stant by adjusting the rates of flow of air and water. The heat 

EMITTED ALONG bd IS THE HEAT THAT WAS ABSORBED ALONG, cd 
(plus something). Then expand the gas adiabatically along ac. 
the temperature falling and the initial condition being restored, 

We have now done work on the gas, or expended power. 
For what purpose? In order that heat might be taken up by 
the gas at 0° and discharged by it at 150°. What good object 
does this serve? 

Suppose it is desired to make ice. Water is available at 50°. 
Our gas, at 0°, can cool and freeze this water. For the opera¬ 
tion to continue, it must then get rid of the heat which it has 
taken from the water. By expending power we raise the gas 
temperature without adding heat and the heat taken from 
the water frozen can then be transferred to the circulating water 
in the coils of the “condenser” and so discharged from the system. 

This is the operation of a refrigerating machine. In the 
Carnot cycle used for refrigeration, all changes of temperature 
would take place without any flow of heat, and all transfers of 
heat would take place without changes of temperature. Actual 
refrigerating machines do not realize these conditions. 

This is unfortunate. The efficiency of the Carnot cycle used 
for refrigeration is 


mcdM _ T 2 
mabM — mcdM 1\ — T 2 ’ 


(25) 


A reversal of the argument of Art. 34 shows that this is the 
maximum possible value. No actual refrigerating machine can 
surpass this efficiency. 

Efficiency is effect divided by cause. The effect, in refrigera- 




Permanent Gases — Laws and Processes 43 

tion, is the absorption of heat by the fluid at low temperature. 
The cause is the mechanical work done. The part played by 
the circulating water being disregarded, as in Equation (25), 
the efficiency may exceed 100 per cent. It is a maximum when 
the temperature range is least. With engines, on the other 
hand, efficiency is a maximum when the temperature range is 
greatest. 

Prob. 79. What would be the percentage saving of coal, in 
Prob. 77, if the exhaust temperature could be reduced to 40° F.? 

(Ans., 33^ per cent.) 

Prob. 80. With the gas temperatures specified in Art. 35, 
what is the efficiency of refrigeration in the Carnot cycle? 
(Ans., 3.07 or 307 per cent.) What would it be if heat were 
rejected at 50° F.? (Ans., 920 per cent.) 

Prob. 81. Under the gas temperatures specified in Art. 35, 
the heat absorbed by the gas (useful refrigeration) is 307 000 
B.t.u. per hr. What amount of mechanical work must be done 
on the gas? (Ans., 100 000 B.t.u. per hr.) What engine horse 
power is ideally required? (Ans., 39.3.) 

36. Availability of Heat. Recapitulating: for efficient power 
generation in a Carnot cycle, the heat should be supplied at a 
high temperature and discharged at a low temperature. The 
latter temperature should be low, but is more or less fixed by 
surrounding conditions. The supply temperature is the one 
which is controllable. It should be high for high efficiency of 
the Carnot cycle. No actual engine can exceed the efficiency 
of the Carnot cycle. Anything that keeps the latter efficiency 
low is bound to keep the former efficiency low, also. Hence 
for all engines, a high supply temperature is desirable. Any¬ 
thing that lowers temperature lowers the potential efficiency. 
A fall of temperature implies a loss of possible power. The 
availability of heat is determined by its temperature. A foot 
pound is 1/778 of a heat unit: but a much smaller fraction of a 
heat unit is actually realized in mechanical work by actual 


44 


Thermodynamics, Abridged 


engines. Low fluid supply temperatures make the fraction 
particularly small. Steam at 600° F. is worth potentially about 
twice as much as steam at 300° F. It does not cost twice, as 
much. 

Whenever the temperature falls without the simultaneous 
performance of mechanical work there is a loss of availability 
of the heat which can never be recovered. This principle is 
sometimes referred to as the second law of Thermodynamics. 


37. Carnot Cycle Impracticable. An engine should be efficient 
and powerful. The latter condition renders it necessary that 
the mean effective pressure, p m , should be fairly high. This 
should be accomplished without excessive maximum pressures. 
In the Carnot cycle, Fig. 13, 



nearly, since for any perfect gas y is approximately 1.4. Thus 
if T i is twice T 2 (which would give an efficiency of 0.50), P b 
would be 11.3Pd. P a is still greater. Suppose we narrow the 
Carnot cycle by moving ac toward bd. Ultimately the two 
curves coincide, the enclosed area becomes zero, p m = 0 and still 
the maximum pressure is 11.3 times the lowest pressure for the 
temperature range stated. 

For a finite area, the mean effective pressure is 


area _ 2 IF / V b 

v d - v a v d - v a ~ f pJ a log€ v a + 


Vb p b V b — p d v d 


T t- , Vd PaV a — PcV ( 

- VcVc \oge-r7 --- 

y c y 


y - i 

I [—)- (V d — V a ) 


J / T7 j7 \ 1 I b paf a 

= | (PaV a - Pc V c ) l0ge-+ ——[- 


+ V —y _ Y~ ~ } + (V* - Va). 

Since p c V c = PdV d and p b V b = p a V a , the last two terms within 
the bracket disappear. Then 








Permanent Gases — Laws and Processes 


45 


(Ta - T c )wR log , 22 

_ __ Pb 

' ) 

\ Pd Pa) 


(Ti-Tt) log, 



. (26) 


Pd Pa 


For P./Pi = 0.5, y = 1.4, p d = 15, p a = 300, T l = 1000, this 


gives 

= 500 X 2.3 log (20 X 0.5 3 - 5 ) 
Pm ~ 500 _ WOO 

15 300 


9.5 lb. per sq. in. 


Thus, although the efficiency is 50 per cent., the mean effective 
pressure is only 9| lb. per sq. in., in spite of the fact that the 
maximum pressure is 300 lb. per sq. in. This would never do. 
It would require a very large cylinder in order to develop even a 
small amount of power. A steam engine, not working in the 
Carnot cycle, might easily give p m = 100 when the maximum 
pressure was 300 lb. It would thus be more than ten times as 
powerful as the Carnot engine. This would compensate for its 
lower efficiency. 

Prob. 82. A Carnot cycle has an efficiency of 0.4. Find the 
lowest possible value of PJPd, using y = 1.4. (Ans., 5.98.) 
What is then the value of p m ? (Ans., 0.) 





CHAPTER III 


AIR ENGINES, AIR COMPRESSORS, 
AIR REFRIGERATION 


Ideal Cycle 

38. Joule Cycle. A cycle underlying the operation of several 
types of air machinery is shown in Fig. 14, diagram abed. It is 

bounded by a pair of adia- 
batics and a pair of constant- 
pressure lines (hence by two 
pairs of isodiabatics — Art. 
31). The efficiency is 
HH/Hab, H dc being negative 
and Hda&nd H bc being both 
zero (compare Art. 34). 



H a , 


H cd 


= 1 - 


H ab 

Hcd 

H cib 

kw(T c — T d ) 


1 kw(T b — T a ) ’ 

where w = weight of fluid 
and k = specific heat at constant pressure. From Equation (20), 


r rn rii rji rr rp rp 

_£ _ h h _ 1 = _i. b _ 1 H - i d Jj 

m rf ) rn rii -*•, m rji rp j 

id la Id la l b — 1 a la 


T d T a T d 

T a T 


This looks like the expression for efficiency of the Carnot cycle 
(Art. 34), which is, however, in terms of the limiting tempera¬ 
tures. For Fig. 14, the Carnot efficiency is {T b — Td)/ T b , 
which is higher than the Joule efficiency. 

46 














Air Engines, Air Compressors, Air Refrigeration 47 


The compression ratio (ratio of pressures during the adiabatic 
compression da which closes the cycle) being called C, we have 


C 


Pa 

P d 



vl(v- 1 ) 


6=1- C a ~ y) 'v. 


(28) 


Prob. 83. Find the efficiency of a Joule cycle with a com¬ 
pression ratio of 10 when y = 1.406. (Ans., 0.486.) If the 
lowest temperature is 500° abs., and y = 1.406 what is the 
temperature at the end of adiabatic compression? (Ans., 972° 
abs.) 

Prob. 84. If T b , Fig. 14, is 1500° abs., what is the Carnot 
efficiency in the foregoing problem? (Ans., 0.667.) 

39. Mean Effective Pressure. The exponent y of the curves 
be and ad is simply a special value of the polytropic exponent n. 
The mean effective pressure (average ordinate) is 


Pm = 


abed 

/T T T 7 \ I PbPb PcPc /T f jy ^ 

VaiPb-Pa) H-[- Pc(Pc-Pd) 

V c - V a 


PaPg—pdPd 
n — 1 


(Arts. 16, 26) 


= —j ( p b Pb — p a Pa — PcPc + PdiPd) (Pc — Pa) 

71 I 


= — {MVi - Va) + Pd(V d - V c )\ -s- (V c - V a ). 

71 I 

Now clearance = c = V a + (P c — P a ): hence ( V c — P a ) 
= Va/c. Also Pb/pd = C. 

enpd C(Vb — Pa) + Pd — Pc 

Vm = rT^l * V7~ 

_ enpd ((jPb _Pd_ F c \ 

Tl 1 \ Pa Pa Pa J 


Substitute 











48 


Thermodynamics, Abridged 


Y*-mn Zi-t+1 c-i/». c -+i 

V a -°’V a ~c’V a V c Va C ‘ 

C (n-l),n . C _±l - C + C l ' n ~ —~ ) 
c c ) 

= ((c + - 1) + c(C lln - C) ). (29) 

This holds whether n = y or not. 

Prob. 85. In Fig. 14, pa = normal atmospheric pressure, 
C = 10, T d = 500, T b = 1500, n = 1.4. Find T a (Ans., 965), 
T c (Ans., 778), V c /V d (Ans., 1.56), V d /V a (Ans., 5.18), V c /V a 
(Ans., 8.05), c (Ans., 0.142), p m (Ans., 19.6). 

Compressed Air 

40. Air Machinery. This section deals only with those air 
machines in which the pressure changes are of sufficient magni¬ 
tude to introduce noticeable 
temperature effects. It ex¬ 
cludes such devices as cen¬ 
trifugal fans and blowers. It 
considers the air as pure and 
dry and treats the fluid as a 
perfect gas. Properties of 
moist air are to be considered 
later. 

41. Compressor. A com¬ 
pressor is the reverse of an 
engine. Work is done on 
the fluid (air) by the supply 
of power from a steam cylin¬ 
der, an electric motor, or 
some other suitable source. Fig. 15 represents the essential 
parts of a compressor cylinder C, with piston, inlet valve a, dis¬ 
charge valve b and receiver D. Suppose the piston to be at the 
extreme left, and both valves to be closed. It is moved toward 


























Air Engines, Air Compressors, Air Refrigeration 49 

the right. At a suitable point early in this stroke, the inlet valve 
is caused to open. The piston suction draws air into the cylinder 
throughout the rest of the stroke. At the end of the stroke, the 
inlet valve is closed. The piston now moves to the left, com¬ 
pressing the air. When the pressure has been sufficiently in¬ 
creased, the discharge valve opens. The high-pressure air is 
then discharged to the receiver and pipe line throughout the re¬ 
mainder of this left-hand stroke. When the stroke is completed, 
the discharge valve is closed. The air left in the clearance space 
expands, during the first part of the succeeding right-hand 
stroke. When its pressure has been reduced to that of the at¬ 
mosphere, the inlet valve opens and the cycle is repeated. 

Practically all compressors are double-acting. 

In Fig. 14, the inlet valve opens at d. Air is drawn in along dc. 
The pressure along dc is a pound or so below that of the atmos¬ 
phere. The inlet valve closes at c and the air is compressed as 
along cb. If the compression were infinitely rapid, the com¬ 
pression path would be an adiabatic, for there would be no time 
for heat to escape. If compression were infinitely slow, the 
curve would be an isothermal, because any increase of temper- 
ture would be offset by conduction of heat. Power is saved in 
the latter case. The lower the value of n the Less is the area of 
the cycle (which is counter-clockwise) and the less is the 
power consumed. The value of n depends on the piston speed 
and the thoroughness of cooling by the water jacket which sur¬ 
rounds the cylinder. (Some small compressors are air-cooled 
only.) Usually, n is between 1.3 and 1.35. The discharge 
valve opens at 6. The pressure along ba is a pound or so higher 
than the receiver pressure and does not vary much, because of 
the relatively large receiving capacity into which the air is dis¬ 
charged. The discharge valve closes at a. The clearance air 
then expands as along ad, the value of n for this curve being 
practically the same as for cb. 

42. Compressor Power Consumption and Capacity. Equation 
(29) gives a very close approximation to the mean effective 


50 


Thermodynamics, Abridged 


pressure of an actual compressor if a proper value is selected for n. 
The value of pd may be taken in good machines at 14 lb. per 
sq. in. Clearance averages 0.04. C is determined by the dis¬ 
charge pressure desired. If S = piston speed, ft. per min., 
N = r.p.m., d = piston diameter, in., D = displacement, cu. 
ft. per min., the indicated horse power of the compressor cylinder 
(compare Art. 7) is (double-acting) 

_ 7T p m d 2 S _ 144p m 2) 

Ihp. - 4 * 33 ooo “ 33 000 

and the stroke in ft. is S -f- 2 N. Values of S are from 400 to 
800. The primary horse power required to drive the compressor 
is the above power divided by the mechanical efficiency of the 
compressor and its driver. 

The ideal action might be considered as realized when there 
was no clearance, when pd = 14.696 lb. per sq. in., and when 
compression (path eg) was at constant temperature. Then 

D=V c -V e = V c , 

SIT = W eg + W gc - W C f 

= P g (V„ - V e ) + P 0 V g log, f‘ - P C (V C - Vi). 

But Ve = 0, Vf = 0, PgVg = PcVc. 

s W = P C V C log,g= 144 X 14.696 X 2.3026 D lo S( 14 ^ 396 ) ’ 
SIT 

Dip. = 3 - 3 -^ = 0.1482) (log p g - 1.167), (30) 

SIT 

Vm = ^ = 33.84 (log p g - 1.167). (31) 

Here (since D is taken as displacement per minute) Fig. 14 is 
to be regarded as a summation of all cycles realized in one minute 
of time. 

In the ideal cycle, the volume of air drawn into the cylinder 
at atmospheric pressure is V c = 2 ). In an actual cycle the 
volume is always less than 2). Suppose in the cycle abed the 






Air Engines, Air Compressors, Air Refrigeration 51 

line hj to represent normal atmospheric pressure. The whole 
volume of air compressed, measured at this pressure, is Vj. Part 
of this is clearance air; the volume of which, also measured at 
atmospheric pressure, is F*. Then the volume drawn in and 
compressed, measured at atmospheric pressure, is Fy — Vh- 
Free air is air at normal atmospheric pressure. 

Compressor capacity is measured by the volume of free air 
drawn in and compressed per minute'(symbol F). 

Volumetric efficiency (not an efficiency in the thermo¬ 
dynamic sense) is the ratio of free air drawn in and compressed, 
to the displacement. 

e* = F/D. (32) 

For the ideal cycle fcge, Fig. 14, F = D and e v = 1.0. For the 
actual cycle dcba, 

F 

e v- D - 
for, 

V c -V a 


Pa 

14.696 

High volumetric efficiency (which leads to large capacity for a 
given size and speed), results when the clearance is low, the 
valve ports and passages of ample area, the piston speed low 
and the value of n low. It is usually between 0.70 and 0.90. 

Prob. 86. An air compressor of 2000 cu. ft. capacity has an 
intake pressure of 13.226 lb. per sq. in., absolute, 4 per cent, 
clearance, a discharge pressure of 132.26 lb. gauge. Find p m if 
n = 1.35. (Ans., 35.2 lb.) 

Prob. 87. In the previous problem, what is the volumetric 
efficiency? (Ans., 0.742.) 

Prob. 88. In the foregoing, find displacement and Ihp. 
(Ans., 2700 cu. ft. per min., 415 hp.) 












52 


Thermodynamics, Abridged 


Prob. 89. The compressor cylinder is directly driven by a 
steam cylinder and the steam cylinder develops 532 Ihp. What 
is the mechanical efficiency? (Ans., 0.78.) 

Prob. 90. What would be the displacement of an ideal com¬ 
pressor giving the diagram fcge, Fig. 14? (Ans., 2000 cu. ft. per 
min.) What would be its Ihp.? (Ans., 296.) 

Prob. 91. In the actual compressor (Prob. 88 ), the piston 
speed is 800 ft. per min. and the rotative speed 100 r.p.m. Find 
cylinder diameter and piston stroke. 

(Ans., diameter = 24.9 in., stroke = 4_ft.) 


43. Cooling. Low values of n for the compression curve are 
realized as a result of jacket cooling. The limit of gain arises 
from the slow rate of heat transfer from air through the cast 
iron walls of the cylinder (Chapter VII). Below this limit, the 
heat lost by the air, in B.t.u. per min. is weight X specific 

HEAT X TEMPERATURE CHANGE 

w a s a {T c - T b ) 


| P e (l + c)D 
[ 53.36 T c 


10.1689 


n - 1.406 
n — 1 



This may be equated to the heat gained by the water 

= w w (t 2 — h), 


in all of which w a and w w are weights of air compressed and 
water. circulated, lb. per min., s a = specific heat of air for a 
polytropic path having the exponent n, t 2 and ti are outlet and 
inlet temperatures of water at the jacket. Normal values are 
T c = 660 (it cancels out anyway), t 2 = 130, t\ = 60 or 70. 

Prob. 92. In the problems of Art. 42, find s a , w a and w w if 
T c = 660, h ■= 70°, t 2 = 130°. (Ans., = - 0.027, w a = 152, 
w w = 39.) What is the temperature at 6? (Ans., 767° F') 

44. Two-Stage Compressor. Compression in successive stages 
saves power. In Fig. 16, compression begins at c. ct is an 
adiabatic, cz a poly tropic, cs an isothermal. The last is the 




Air Engines, Air Compressors, Air Refrigeration 53 

preferred process, but cannot be realized because heat can be 
transmitted only slowly through the cylinder walls. 

Compress from c to u adiabatically, or slightly better than 
that if possible, then discharge the air to an external cooler which 
reduces its temperature to v. Again compress and cool (vw, wx). 
With an infinite number of steps and complete cooling at each 



step, the isothermal path is realized, and power is saved. A 
number of cylinders and intercoolers are required for multi¬ 
stage compression. The intercooler is a cylindrical shell filled 
with small tubes through which cold water circulates, while the 
air surrounds the tubes, on its way from one cylinder to the next. 

The diagrams abed and ABCD show the action of a two-stage 
machine, with polytropic compression in each cylinder. With 
good intercooling, t c = t c . The gross saving of work, as com- 










54 


Thermodynamics, Abridged 


pared with single stage compression, is represented by the area 
Bzbr. To offset this, partially, there is half the duplicated 
hatched area due to friction of air passing through the intercooler, 
which makes the suction pressure of the second cylinder a pound 
or so less than the discharge pressure of the first, and thus some¬ 
what decreases the saving. 

Multi-stage operation saves power, though increasing first 
cost and complication. The saving is greatest if the discharge 
pressure is high. The higher this pressure, the greater the 
number of stages desirable. Two stages are common in shop 
practice, with discharge pressures around 100 lb. gauge. 

45. Power Required in Two-Stage Compression. The least 

total power is consumed when the intercooler pressure is 

Po = ViVV 

The two cylinders will then (if tc — t c ) require equal amounts 
of power. The suction pressure pc, with properly designed valve 
passages, may be taken at p 0 — 0.5, and the discharge pressure, 
p a , at po + 0.5. Equation (29) gives the mean effective pressure 
of either cylinder, but the following points should be observed: 

First (low-pressure) cylinder, clearance = Ci = v a l(v c — v a ), 

C = Pa/Pd- 

Second (high-pressure) cylinder, clearance = Ch = va/(vc—Va), 
C — Pa/pd, Pd should read p D . The equations of Art. 42 may 
then be used to obtain the Ihp. of each cylinder, for a given 
displacement, D, of the Low-pressure cylinder. With perfect 
intercooling, as shown in Fig. 16, the high-pressure displace¬ 
ment is D X Pclpc- 

Thus, take the conditions of Prob. 86. 

po = V13.226 X 146.96 = 44.1. 

Then pc = 43.6, p a = 44.6. For the first cylinder, the mean 
effective pressure is 

13.226w f , f / 44.6\ (n ~ 1)/w 1 

Vmi ~ v^r[ (Ci + 1) i vi3^j -1 | 




Air Engines, Air Compressors, Air Refrigeration 55 


For the second cylinder, it is 

43.6ti f f / 147.0 \ 

+ Ch 


(n—l)/n 


- 1 


/147.0 V' n _ ( 147.0 \ [1 
\ 43.6 / \ 43.6 y J J * 


The displacement of the second cylinder is 

The horse power consumed in the two cylinders is 


144 Di / 13.2 \ 

33 000 V Vml + 43.6 pml ) ' 


46. Capacity, Two-Stage. The volumetric efficiency e v i of the 
first cylinder is given by Equation (33) and Di = F - f- e v i. It is 
higher than the volumetric efficiency of the single-stage machine 
with the same extreme pressure limits ( Pa and p c ) and the same 
clearance. 

Prob. 93. In Art. 45, with n = 1.35, Ci = Ch = 0.04, find 
Pmi and p mh . (Ans., p m i = 18, p mh = 58.) 

Prob. 94. In the foregoing, what is the ratio of displacements 
of the two cylinders? (Ans., DJD H = 3.3.) 

Prob. 95. In the foregoing, what is the volumetric efficiency 
of the first cylinder? Compare Prob. 87. (Ans., 0.87.) 

Prob. 96. If F = 2000, E vi = 0.87, find D L , D H , in the fore¬ 
going. (Ans., D l = 2300, D H = 700.) 

Prob. 97. Find the horse power consumed by the whole 
machine, noting that the cylinders require practically equal 
powers. (Ans., 356 hp.) How much power is saved over single 
stage operation as computed in Prob. 88? (Ans., 59 hp.) 

Prob. 98. With S = 800, r.p.m. = 100, state cylinder dimen¬ 
sions. (Ans., both strokes are 4 ft.: diameters are 23 and 12.65 
in.) 

5 




56 


Thermodynamics, Abridged 


47. Heat Removed in Two-Stage Compression. The method 

of calculation is much the same as for single-stage machines* 
given in Art. 43. In Art. 45, and Probs. 93-98, assume T c = 660. 
Then T b = T c { Vb lv c Y n ~ 1)ln = 660(44.6/13.226) 0 - 26 = 904. Assume 
T c = 660, as should easily be the case with reasonably good 
intercooling. Then T B = T c{'PBlvcY n ~ V),n = 904, also. The air 
temperature is reduced 904 — 660 = 244 degrees in the inter¬ 
cooler. It is increased 244° in each cylinder. The weight of 
air in the low-pressure cylinder is P c (l + Cj)Di -f- 53.36 T c 
= (13.226 X 144 X 1.04 X 2300) 4- (53.36 X 660) = 130 lb. per 
min. The weight in the second cylinder is practically the same 
if its clearance is the same. The weight in the intercooler may 
differ somewhat, since the intercooler does not cool the clearance 
air: this difference will be disregarded. The passage of air 
through the intercooler may be regarded as taking place at 
constant pressure. The specific heat there is consequently 
k = 0.2375. The specific heat of air during compression in the 


cylinders is 


s — 


n — 1 


0.1689 X 


- 0.056 
0.35 


0.027. 


Then the heat removed from the air is its weight multiplied by 
the sum of the products of specific heats and temperature ranges, 
or 

130{ (0.2375 X 244) + (2 X 0.027 X 244)} = 9250 B.t.u. per min. 
This must be removed by water supplied at the jackets and 
intercooler. It considerably exceeds the heat removed in single- 
stage compression, where there is no intercooler. 

Prob. 99. How much heat was removed in Prob. 92? 

(Ans., 2330 B.t.u. per min.) 

Prob. 100. In Art. 47, 150.4 lb. of water are delivered to the 
intercooler every minute. How much will be the rise of tempera¬ 
ture of this water? (Ans., 50°.) 

Prob. 101. In Art. 47, the cooling water enters the jackets 
at 70° and leaves at 156.5°. How much water must be supplied 
at each jacket? (Ans., 9.9 lb. per min.) 




Air Engines, Air Compressors, Air Refrigeration 57 

48. Multi-Stage Compression in General. For four stages, 
the intercooler pressures are determined by an extension of the 
method of Art. 45. There will be three intercoolers. The 
pressure in the middle one, P 3 , will be a mean proportional 
between the suction pressure, Pi, of the lowest pressure cylinder 
and the discharge pressure, P 5 , of the highest pressure cylinder. 
The pressure in the first intercooler, P 2 , will be VPiP 3 . That 
in the third will be P 4 = VP 3 P 5 . The method may be further 
extended for eight stages. For three stages, with successive 
suction pressures of Pi, P 2 and P 3 and a final discharge pressure 
of P 4 , the condition for least power consumption and (with 
intercooling sufficient to restore the pre-compression temperature 
in each case) for equal power consumption by all three cylinders is 

P 3 = a/PJY, P 2 = ViVY. 

Prob. 102. A four-stage compressor operates between pres¬ 
sures of 14 and 250 lb. absolute. What are the best intercooler 
pressures? (Ans., 29, 59, 122.) 

Prob. 103. What would be the best intercooler pressures in a 
three-stage compressor for the same conditions? 

(Ans., 37 and 96 lb.) 

49. Vacuum Pump. The vacuum pump connected with a 
steam engine or turbine condenser is an air compressor having 
a suction pressure between 0.5 and 2.0 lb. absolute and a dis¬ 
charge pressure around 15 or 16 lb. absolute. Because of the 
presence of water in the cylinder, the value of n is low. 

50. Air Engine. The most common example of an engine 
using compressed air is the pneumatic tool. Suppose some of 
the air compressed, in Prob. 86, to be used in an engine. The 
essential parts of an engine are shown in Fig. 17, the upper 
diagram indicating the cylinder, C; piston, P; piston-rod, R; 
inlet passage, /, from the compressor; and outlet passage, E , 
to the atmosphere. These passages have valves, not shown. 

The piston is at the left-hand end of its stroke, position Pi. 
The inlet valve is open. Hence there is a pressure on the piston, 






58 


Thermodynamics, Abridged 


tending to force it toward the right. The compressor discharge 
pressure (Prob. 86) was 146.96 lb. absolute. The pressure in 
the engine cylinder will be somewhat less: we will take it at 
145.53 lb., as indicated. In the lower diagram, the point a 
represents the state of the fluid, V a being the clearance volume. 



The piston moves toward the right, until the position P 2 is 
reached. The inlet valve is then closed (point b on the lower 
diagram). The piston will continue to move to the right, 
impelled by the expansive force of the air, giving the process 
curve be. The equation of this curve is jpbVb n = 'p c V c n i where n 
is usually around 1.3 to 1.35. High speed or a heat-insulated 
condition of the cylinder tend to increase the value of n and thus 
to decrease work and power. The temperature falls along be. 

Expansion terminates at c, when the discharge valve opens, 
air rushes out of the cylinder through the outlet E and the 
























Air Engines, Air Compressors, Air Refrigeration 59 

pressure falls, the piston having reached the end of its stroke 
(path cd). The farther the expansion is carried, the more work 
will be obtained from the expansion of a given quantity of air. 
On the other hand, long expansion leads to low mean effective 
pressure and hence to large size of engine for a given power. We 
will assume cut-off (the closing of the inlet valve at b) to occur 
at one-quarter stroke. (The diagram is not drawn to scale.) 

Now on a volume scale, Vd — V a = D, the displacement per 
stroke. If clearance be taken at 7 per cent., V a = 0.07D. If 
cut-off is at quarter stroke, ab = D/4: or 0.25D and Vb = 0.25 D 
+ 0.07 D = 0.32 D. The pressure at c is pb(Vb/V c ) n • With 
n = 1.3, this gives p c = 145.53(0.32/1.07) 1 * 3 = 30.3. 

The piston is now in the position P 3 . It moves to the left 
(by the inertia of the reciprocating parts, or if the engine is 
double-acting, under the action also of high-pressure air on its 
right-hand side). Throughout the first part of the left-hand 
return stroke, the pressure remains about constant and a little 
above that of the atmosphere. (It must be somewhat above, 
else air would not flow out of the cylinder.) We will take the 
pressure at 16.17 lb. 

At some position P 4 , before the return stroke is completed, the 
outlet valve is closed. The comparatively small amount of air 
then remaining in the cylinder is compressed (path ef) during the 
remainder of the return stroke. The point e, at which the exhaust 
valve closes is called the point of compression (strictly/ “ be¬ 
ginning ” of compression). The object of compression is to 
cushion the action of the engine. We will assume compression 
to begin at 8/10 the return stroke. Hence F e = 0.27D. 

The path ef follows the law p/V / w = p e V e n , the value of n 
being again about 1.3. Then 

The compression curve terminates at /, the piston having then 
again reached its extreme left-hand position. The inlet valve 
immediately opens and the pressure increases, giving the path fa 
before the piston starts to move again. 


60 


Thermodynamics, Abridged 


51. Air Engine Design. Still referring to Fig. 17, the work 
done in the cylinder by the air is, since Wf a = 0 and W cd = 0, 

2 JF = W ab + W bc - W de - W ef 

PbV b — p c V c 


144 2) 


j MVi, - v a . 


) + 


n — 1 

- M Vt - V.) - 


PjVf - PeVe 

n — 1 


= 1442) (145.53 X 0.25) + 


(145.53X0.32)-(30.3X1.07) 
0.3 


- (16.17 X 0.8) - 


(93.5 X 0.07) - (16.17 X 0.27) 
0.3 


double- 


= 91202) ft. lb. per revolution, on each side of 
acting piston. 

Suppose 10 hp. to be required, the piston speed to be 300 ft. 
per min. and the piston diameter to equal the stroke. If N 
= r.p.m., S = piston speed, d = dia. in in., L = stroke in ft., 

d 3 d 3 

9120 D X 2JV = 10 X 33 000, " " 


D = - • 

4 1728 


2200 ’ 




iV 2 L d ’ 

330 000 X 2200 
9120 X 12 X 300 : 


d 3 


12 S 

9120 X ioo x-5- = 330 00 °- 

4.7 in., L = 4.7 -M2 = 0.392 ft., 
4.7 s 


N = 1800 4- 4.7 = 383 r.p.m., D = = 0.0472 cu. ft. 


52. Air Consumption. The volume of fresh air present at b, 
Fig. 17, when the supply is cut off, is V b = 0.322). Some of this 
is clearance air. How much? Not V a , for some of the air at a 
is fresh air, admitted during the operation fa. We know that 
the volume of clearance air alone was V a == F/, at the pressure 
Pf. What is it at the pressure p 0 ? It is reasonable to assume a 
continued compression of clearance air by incoming air, along 
the path fx, which is the path ef produced. Then the volume 
V x is that of clearance air present at the pressure p a , when the 
total volume of air received is V b . The value of V x is 










Air Engines, Air Compressors, Air Refrigeration 61 


Hence the volume of fresh air supplied to the cylinder by the 
compressor is V b — V x = 0.27 D = 0.27 X 0.0472 = 0.0127 cu. 
ft. per stroke or 0.0127 X 383 X 2 = 9.7 cu. ft. per min. This 
is air at 145.53 lb. pressure. Measured as free air, its volume 
would be 9.7 X 145.53/14.696 = 96 cu. ft. per min. 

The air rate, which measures the economy of the engine, is 
the VOLUME OF FREE AIR USEJ) PER HOUR PER HP. Its Value for 
our conditions is 


96 X 60 

10 


576 cu. ft. 


53. Comparison with Ideal Conditions. The ideal air engine 
cycle would have no clearance, expansion would be “complete” 
(i.e., cut-off would occur so early that the points c and d would 
coincide), the exhaust pressure would be atmospheric and n would 
equal 1. Cycle egcf, Fig. 14, will represent the action if p/ c 
= 14.696. Then the fresh air = eg, free air = eg X ValVc — f c > 
and the work of the cycle is 

Pe(V g - V e) + P 0 V e ] 0ge P - P,(Vc - V S ) 

v o 

= 144 X 14.696F X 2.3 log = mOF log 

F denoting the volume of free air used to produce this work. 
If s cycles are produced per hr., the free air consumption is Fs 
cu. ft. per hr. and the hp. is 

4860F* p g 

60 X 33 000 l0g 14.696’ 

so that the ideal air rate is 

Fs_ _ 1 980 000 = 407 

hp. “ 4860 (log p g — 1.167) lo gp g - 1.167 * 

For our conditions, p g = 145.53, log p g = 2.163, ideal air rate 
= 407/0.996 = 409 cu. ft. The “relative efficiency” of our 
engine, or efficiency as compared with that attainable under 
ideal conditions, is 409 -f- 576 = 0.71. (This is not a true 
efficiency, but only the ratio of two efficiencies.) 









62 


Thermodynamics, Abridged 


The mean effective pressure of our actual engine is 

W 9120 n „ 

Vm = D = 144 = 63,3 b * per Sq ‘ m * 

That of the ideal engine is given by Equation (31) and is for our 
conditions 33.6 lb. per sq. in. If both engines have the same 
dimensions and speed, the powers are in the same ratios as the 
mean effective pressures. The actual engine develops 10 hp. 
The ideal engine will therefore develop only 

33 6 

X 10 = 5.3 hp. 

Its lower mean effective pressure is due to its “complete” expan¬ 
sion, which makes the pV diagram depart more noticeably from 
a rectangle. 

54. Preheat. Reference has been made to the fall of tempera¬ 
ture which occurs along be, Fig. 17. This is important. The 
air may cool below 32° F. Any moisture then present congeals 
and may obstruct the action of the engine. This condition may 
be avoided by using very little expansion, as in pneumatic tools. 
Another method is to heat the air, close to the engine, in a sort 
of hot air furnace or preheater. An interesting result follows 
the use of the preheater. 

Assume in Fig. 17 that T b = 560° and that by employing a 
preheater this temperature can be raised to 840°. Then, with¬ 
out any change in the quantity of air admitted to the cylinder, 
its volume at cut-off becomes 

T 840 

V ° =Vb f b = °* 322) X 565 = 0A8D > 

by Charles’ law. The air now expands along gh, the law being 
such that 

ph = p 0 {n) n= 145 - 53 (of ) 1 * = 51 - 6 - 

Added work is gained, equal to the area bghc, the value of which is 

w bg + w gh - w bc 


Air Engines, Air Compressors, Air Refrigeration 63 


= p g (y g - v b ) 


P 0 V g - PkVa 

n — 1 


P b V b - P C V C 

n — 1 


= 144 D 


(145.53 X 0.16) 


(145.53 X 0.48) - (51.6 X 1.07) 
h 0.3 

(145.53 X 0.32) - (30.3 X 1.07) 
0.3 


= 144 X 0.0472 X 26.3 = 178 ft. lb. per cycle. 

This is equivalent (there being 766 cycles per min.) to 4.1 hp., 
ADDED TO THE ENGINE OUTPUT WITHOUT ANY INCREASE IN SIZE 
OR AIR CONSUMPTION. 

What does this cost? The preheater must supply enough 
heat to warm all the air in the cylinder (both fresh air and 
clearance air) from 560° to 840° absolute. The specific heat may 
be taken as at constant pressure, h = 0.2375. The weight of 
air is P b V b -f- RT b = (144 X 145.53 X 0.32 X 0.0472) (53.36 

X 560) = 0.01056 lb. per cycle. The preheat per cycle is then 
0.01056 X 0.2375 X 280 = 0.7 B.t.u. or 545 ft. lb., and the 
efficiency of the preheating operation is 178 -r- 545 = 0.326. 
The importance of this will now be shown. 

55. Efficiency of Compressed Air Plant. In Probs. 86 and 88, 
it was found that a 2000 cu. ft. (free air capacity per minute) 
single-stage compressor discharging at 132.26 lb. gauge pressure 
(146.96 lb. absolute) required 415 indicated hp. in the air cylinder. 
Prob. 89 gives the corresponding hp. of the steam cylinder which 
drives the compressor as 532. Prob. 90 shows that the ideal 
compressor would have required only 296 hp., measured in its 
own cylinder. 

If we start with 532 Ihp. steam, we lose at once 532 — 415 
= 117 hp., in friction of compressor mechanism. We then lose 
415 — 296 = 119 hp. in the compressor cylinder due to departure 
from ideal conditions. The “ relative efficiency ” or ratio of 
actual and ideal efficiencies of this cylinder is 296 -r- 415 = 0.713. 

The compressed air reaches the engine (Art. 52) at a pressure 
somewhat less (145.53 lb. absolute) than that at which it was 
discharged from the compressor, and also at a much reduced 






64 


Thermodynamics, Abridged 


temperature. Having defined the ideal engine and compressor 
cycles as we have, the only loss chargeable to pipe line trans¬ 
mission is the pressure loss, which may be estimated as follows. 

Art. 53 shows that the ideal air rate for our engine is 409 cu. ft. 
of free air per Ihp. hr. If we installed a sufficient number of 
engines to use all the air furnished by the compressor, the ideal 
Ihp. of all of these air engine cylinders would be (2000 X 60) 
-r- 409 = 293. The pipe line loss is then 296 — 293 = 3 hp., 
or about 1 per cent. If the compressor discharge and engine 
inlet pressures had been the same, this loss would have been zero. 

By Art. 52, the air rate of the actual engine is 576. Hence 
the total Ihp. available in actual engine cylinders is (2000 X 60) 
-T- 576 = 208. The “ relative ” efficiency of our engine is as 
already stated, 208 -5- 293 = 0.71. 

The efficiency of the whole conversion process, from Ihp. of 
steam engine to Ihp. of air engine, is then 208 -f- 532 = 0.391. 
This is not a thermodynamic efficiency covering the conversion 
of heat to work, but merely an efficiency of conversion from 
work in one form to work in another form. A compressed air 
plant is merely a method for power distribution. A further loss 
occurs in converting the Ihp. of the air engine cylinder to shaft 
hp., this being due to mechanical friction of the air engine. The 
thermodynamic process would start with the heat in the steam 
entering the steam cylinder of the engine which drives the 
compressor. As the efficiency of the steam engine under average 
conditions is around 0.10, the efficiency from steam to Ihp. 
of air engine is about 0.10 X 0.391 = 0.0391: less than 4 per 
cent. 

The preheater “ boosts ” the air engine output to a moderate 
extent, and the efficiency of its part of the process has been 
shown, in one illustrative case, to be 32.6 per cent. Thus its 
effect is to improve the efficiency of the whole system, although 
it is not installed primarily for that purpose. Compressed air 
is unique among transmission systems, in that supplementary 
energy can be supplied at the receiving end with highly beneficial 
influence on efficiency. 


Am Engines, Am Compressors, Am Refrigeration 65 


56. Tests. Observed air rates in well-designed engines are 
from 477 to 1000 cu. ft. of free air per hour per brake hp. Small 
engines without expansion use around 2000 cu. ft. 

The probable fall of pressure in the pipe line may be estimated 
from Unwin’s formula: 



(35) 


£>2 and pi are pressures at end and beginning of line, lb. per 
sq. in., absolute, 

/ = 0.00435 for 6 in. pipe, 0.004 for 8 in. pipe and 0.003 for 
pipe 12 in. or larger, 
u = velocity of air in pipe, ft. per sec. 



G = length of pipe line, ft., 

T = absolute temperature of air, 
d = internal diameter of pipe, ft. 

Prob. 104. A 12-in. pipe line 5000 ft. long carries 10 000 cu. 
ft. of free air per min., the initial pressure being 147 lb. per sq. in.. 
Find the volume and velocity per sec. of air flowing through the 


pipe. (Ans., vol. = 16.67 cu. ft., u = 21.25 ft.) 

Prob. 105. If the temperature of this air is 40° F., find the 
pressure at the end of the pipe line. (Ans., 145 lb. per sq. in.) 
Prob. 106. If T g , Fig. 17, is 840°, find t h - (Ans., 201° F.) 


Air Refrigeration 


57. Joule Cycle Reversed. The fall of temperature which 
occurs during expansion in the cylinder of an air engine may be 
(and has been) employed for refrigeration. In Fig. 14, suppose 
the Joule cycle abed to be worked counter clockwise. Heat is 
absorbed from d to c, the fluid is compressed (the temperature 
rising) from c to b, heat is emitted from b to a and the fluid is 
expanded (the temperature falling) from a to d. (Compare Art. 
35.) The operation dc is a refrigeration of surrounding bodies. 





66 


Thermodynamics, Abridged 


During this operation, the fluid is circulating through pipes 
placed in a brine tank or room to be cooled. 

Refrigeration being the aim in view, the efficiency must be 
regarded as 

H dc H dc _ k(T c -T d ) 
e 2 W H ia - Hie k(T b - To) - k(T c - T d ) 

T b - Ta- T c + Ti 

Since the curves be and ad are isodiabatics, 


Tb 

T a 


h 

T d ’ 


T b - Ta - T c + 




which may of course exceed unity, as in Art. 35. 

The comparable efficiency of the Carnot cycle would be based 
on the range of temperatures during the refrigerating 
process alone, i.e., along dc, and would be T d + (T c — T d ). 
This exceeds the value by Equation (36). 

Prob. 107. In a Joule cycle for refrigeration, the compression 
ratio (C = p a lp d , Fig. 14) is 4. Find the efficiency if y = 1.406. 

(Ans., 2.04.) 

Prob. 108. If in Prob. 107 and Fig. 14, T d = 420, T c = 571, 
find T a , T b and the efficiency of refrigeration in the Carnot 
cycle. (Ans., T a = 628°, T b = 855°, = 2.78.) 

58. Ideal Air Machine. In Fig. 14, the diagram febe may be 
assumed to represent the action of a compressor, eadf that of an 
engine. The area abed represents the difference between the 
power absorbed by the compressor and that generated by the 
engine. In an ice machine, power equivalent to this area, plus 
the power necessary to overcome friction of mechanism, must 












Air Engines, Air Compressors, Air Refrigeration 67 


be provided from a separate source, such as a steam engine. 
Let subscripts c , E , and ^ refer to compressor, air engine and 
steam engine, respectively. Then for 1 lb. of air, 

abed = 778 (H ab - H cd ) = 778 {k(T b - T a ) - k(T c - T d ) }, 

W c — W E abed W c 

W~ ~W^ = W E ~ 1 ’ 

W E = W ea + W ad — Wif = p e (V a — V e ) 

, VaVa ~ VdVi 

H- yiTi - Vi^Yi - Vi) 

= {VaV a - PiVi ) {Ta - Ti), 

Wc _ , , 778 k{T t - Ta - T c + T d ) , 

W E - 1 + Ry{T. - T d ) ' {y ~ 1} ' 

or since 

R = - y (y-l)X 778, 

Wc = 1 + T^T 1 _ T S= V S= D_ C 

W E l 'T a — Td T d V d D e - 

Also, since W s = W c — W E , 

w * 1 )- W ‘{w«- 1 )' 

where D — displacement. 

Prob. 109. Under the conditions of Prob. 108, what is the 
ratio of air engine power to compressor power? Of steam engine 
power to compressor power? (Ans., 0.735, 0.265.) 

59. Ideal Power and Units. The heat absorbed along dc, in 
Fig. 14, for the temperatures assigned in Prob. 108, is 
H = k(T c - T d ) = 0.2375 X 151 = 35.9 B.t.u. per lb. of air. 
The unit of refrigerating capacity is “ice-melting effect. m 
I f water exists at 32° F., then under purely ideal conditions, the 
abstraction of 144 B.t.u. from 1 lb. of this water will convert it 
to ice. Conversely, the addition of 144 B.t.u. to 1 lb. of ice 








68 


Thermodynamics, Abridged 


will liquefy it. The quantity 144 is the latent heat of fusion 
of ice. The “ice-melting effect” of a refrigerating machine, in 
lb., is the number of B.t.u. it absorbs along dc, Fig. 14, divided 
by 144. Suppose the ice-melting effect required is 113 lb. per 
hr. Then the refrigeration to be done per min. is (113 X 144) 
-f- 60 = 271 B.t.u. If this is to be done when, as above, 35.9 
B.t.u. are removed per lb. of air, the weight of air to be circu¬ 
lated per min. is 271 - 5 - 35.9 = 7.54 lb. 

Efficiency is refrigeration divided by the heat equivalent of 
power expended. Its value here (Prob. 107) is 2.04. Hence 
since 1 hp. = 42.42 B.t.u. per min., the steam cylinder hp. 
necessary is (271 -f- 2.04) -r- 42.42 = 3.16 and (from Prob. 109) 
the air cylinder will develop (0.735/0.265) X 3.16 = 8.8 hp. 
while the compressor cylinder will consume 8.8 + 3.16 = 11.96 
hp. In symbols, let Q = B.t.u. of refrigeration done per minute. 
Then steam cylinder hp. = Q -£■ 42.42e, ice-melting effect 
= Q - 5 - 144 lb. per min., lb. of air circulated per min. = Q 
-5- k(T c - T d ) = w a . 

Heat must be removed from the air along ba, Fig. 14, in an 
external “cooler” or “condenser.” The heat removed per min. 
is q = w a k(Tb — T a ). (It should be noted that steam cylinder 
hp. = (q — Q) -T- 42.42, if the curves be and ad are adiabatic.) 

The tonnage of a machine is the ice-melting effect per 24 hr., 
expressed in tons of 2000 lb. If T = tonnage, 

60Q X 24 _ Q 
2000 X 144 200 * 


A unit of economical performance is the ice-melting effect 
PER HR. PER IHP. OF STEAM CYLINDER. Call this I 0 . Then 


_ 60 Q ' Q 
0 144 ■ 42.42e 


17.6e. 


(37) 


Prob. 110. From Prob. 108, find q if 7.54 lb. of air are circu¬ 
lated per min. (Ans., 407 B.t.u.) 

Prob. 111 . What is the Ihp. of the steam cylinder computed 
from q = 407 and Q = 271? (Ans., 3.2.) 



Air Engines, Air Compressors, Air Refrigeration 69 


Prob. 112. For Q — 271, what is the tonnage? (Ans., 1.355.) 
Prob. 113. Find I 0 for the conditions of Art. 59. (Ans., 35.9.) 



60. Dense Air Ice Machine. Fig. 18 refers to the action of an 
actual ice machine, which consists of compressor and air engine 
(expander) cylinders side by side, with a steam cylinder. The 
action of the steam cylinder is not shown by the diagram. The 















70 


Thermodynamics, Abridged 


compressor cycle is EBCG, the pressure limits noted being 
approximately those of practice and the compression ratio 4, as 
used for the Joule cycle, Prob. 107. The engine (expander) 
cycle is HDAF, the pressure differences between GC and HD 
and between AF and EB being somewhat exaggerated. The 
curves are not adiabatic. We will take n = 1.35. The com¬ 
pressor should be jacketed like any compressor. This reduces 
the work of the condenser, which must cool the air from C to D. 
The air engine cycle is given “ complete ” expansion, in order 
that the fall of temperature may be a maximum. 

The air leaving the engine goes to the brine tank or other 
place to be cooled, then returning to the compressor. The 
necessary value of t A is determined by the service. We will 
assume Ia—— 40° F., whence T A = 420. T B will be higher. 
If ice is to be made, it cannot be much above 0° F. (460). It 
will exceed the cold room temperature somewhat, because the 
air will receive heat from the atmosphere on its way from the 
cold room to the compressor. 

In a specific machine, the compressor and expander cylinders 
were 5f by 10 and 4f by 10 inch and connected to the same shaft. 
The ratio of displacements is then D C ID E = (5.75/4.75) 2 = 1.46. 

VaVa VbVb T b VbVb 65 Dc 1 0 _ 

Ta - Tb ’ Ta~ VaVa~ 70' D e ~ 

both clearances being 10 per cent. Then T B = 420 X 1.36 
= 571 or 111° F. This is an unusually high value, but we will 
adopt it, as in Prob. 108. 

Prob. 114. Find refrigeration per lb. of air in Fig. 18. Note: 
consider the pressure as constant during heating from A to B. 

(Ans., 35.9 B.t.u.) 

Prob. 115. Compute T D} T c . (Ans., 582°, 817°.) 

The value of T D (122° F.) is easily realized at the condenser: 
which should in fact be designed to reduce this temperature to 
80°-100° F. The heat to be carried away at the condenser is 
0.2375(Tc — T d ) = 55.9 B.t.u. per lb. of air. 





Air Engines, Air Compressors, Air Refrigeration 71 


The value of D c is (tt/ 4) X (5.75) 2 X (10/1728) = 0.15 cu. ft. 
per stroke. The volume of air drawn into the compressor (and 
therefore circulated through the system) per stroke is Vb — V E 
= 0.82Dc = 0.123 cu. ft. (see Prob. 116). The weight of air 
per stroke is (P B X 0.123) -r- (53.36 T B ) = (65 X 144 X 0.123) 
-T- (53.36 X 571) = 0.0377 lb. Assume N = 100, i.e., 200 single 
strokes per minute, the machine to be double-acting. Then 
the weight of air circulated per min. is w a = 200 X 0.0377 
= 7.54 lb. (Compare Prob. 110.) We then have 

Q = 35.9 X 7.54 = 271 (see Art. 59 and Prob. 114), 
q = 55.9 X 7.54 = 421. 

In this actual engine, the steam cylinder hp. is not determined by 
q — Q, since heat transfers occur along all the curved paths of 
the cycles. The ihp. of each cylinder (compressor and expander) 
is first worked out in detail. Applying Equation (29) to the 
compressor, 

i qj; V 

Pm C = 0.35 U-l^ 0 - 259 - 1) + 0.10(4 0 - 741 - 4)} = 88.6 

and to the engine (expander), 
i « y 70 

Vm E = not- {1.1 (3.5 0 * 259 - 1) + 0.10(3.5 0,741 - 3.5)} = 88. 


The corresponding horse powers are 

2 X 88.6 X 10 X 0.7854 X 5.75 X 5.75 X 100 


Ihp.c = 


12 X 33 000 


11 . 6 . 


( 4 75 \ 2 88 

sTsJ X 8SU3 = 7 ' 8 ' 


(Powers at equal piston speeds are proportional to the products 
of area and p m .) Then the net power to be derived from the 
steam cylinder is 11.6 — 7.8 = 3.8. To this we may add 50 
per cent, to cover friction losses, so that Ihp. s = 5.7. These 
results may be compared with those for the ideal cycle computed 
in Art. 59. It should be noted that compression in the air 
engine is here “ complete/’ i.e., the curve FH, Fig. 18, con¬ 
tinues until the air-supply pressure is attained. 

6 





72 


Thermodynamics, Abridged 


The efficiency is in this case 

_£_ = 2 -Z 1 = ! 13 

42.42 X 5.7 241 

For the ideal cycle, it was 2.04. The tonnage is that of the 
ideal cycle, 1.355. The ice-melting effect per Ihp.-hr. is I o 
= 17.6 X 1.13 = 19.8 lb. The value realized for this depends 
mainly on the temperature range. In practice, the useful value 
of Q is much reduced by radiation, and values of e around 0.5 
or 0.6 or of 7o around 10 are common. 

If the machine were to make ice from water at 100°, it would 
have to cool the water well below the freezing point, first,—say 
to 0°. This would require the extraction of approximately 100 
B.t.u. per lb. Adding the latent heat of fusion, the refrigeration 
necessary per lb. of ice made is 244 B.t.u. The machine con¬ 
sidered could then make a maximum of 271/244 = 1.11 lb. of 
ice per min. During the preliminary cooling, it could lower the 
temperature of 271 lb. of water 1° per min. 

Prob. 116. In Fig. 18, find V E and Vb — V E . 

(Ans., 0.28 D c and 0.82D C .) 

Prob. 117. In Fig. 18, find V c . (Ans., 0.3932)^.) 

Prob. 118. In Art. 60, check the values of p m c and p mE - 
Calculate Ihp. £ by the formula 2 p m LAN -r- 33 000. 

Regenerative Hot-Air Engine Cycles 

61. Remarks. As has been stated, the air engine of a com¬ 
pressed air plant is not a thermal engine but only a part of a 
work-conversion system. Another type of engine, which uses 
high-temperature air, discharging it at a lower temperature, is 
now to be considered. This is really a thermodynamic machine. 
Such engines are called hot-air engines. They are not 
internal combustion engines, for fuel does not enter the cylinder. 
They differ from most engines (not from dense air ice machines) 
in that a fixed mass of fluid is used over and over again. The 
use of the regenerator, common with hot air engines, makes 
the ideal efficiency of their cycle equal to that of the Carnot 



Air Engines, Air Compression, Air Refrigeration 73 


cycle. In this respect they are unique. The regenerator is a 
closed vessel filled with wire gauze, glass or any substance of 
large heat absorbing capacity. 

62. Stirling Cycle. Eig. 19 shows the principle of operation 
of the Stirling engine. The piston being at the extreme left 



position Pi, the state on the lower diagram is d. Valves V 2 
and F 6 are closed, valves Fi and V 5 are open. Hot air from the 
regenerator flows into the cylinder, the temperature rising to T a 
and the pressure increasing as long as the^piston does not move. 
Valve V 5 is now closed and Vq is opened. Heat from the furnace 










































74 


Thermodynamics, Abridged 


causes the air in the cylinder to expand (path ab) and the piston 
moves to the extreme right (position P 2 ). The supply of heat 
is so controlled with relation to the piston speed that the path ab 
is isothermal. The valve V\ is now closed, F 2 and V 4 are opened 
and V 3 closed. Air flows from the cylinder to the regenerator. 
The pressure in the cylinder falls (path be), the piston remaining 
stationary. As the air flows into and through the regenerator, 
its temperature falls: but the inlet temperature at the regenerator 
finally approximates Tb, while its outlet temperature approxi¬ 
mates T c . Valve V 4 is now closed and V 3 opened. Then the 
piston makes a full stroke to the left, forcing air into the corn 
denser. The water supply in the condenser coils is so controlled 
as to withdraw heat without lowering the temperature. The 
path cd is consequently isothermal. 

Under purely ideal conditions, all heat discharged to the 
regenerator along be is regained along da, so that this quantity 
of heat may be disregarded as a debit against the engine. The 
only heat chargeable is H a b, and the efficiency is 


e — 


{flab H c d) • H a b 



T d 


(Art. 34), which is the efficiency of the Carnot cycle. 

The operation of the Ericsson hot-air engine was similar 
and its efficiency is the same, but the regenerative processes 
were at constant pressure instead of at constant volume. 

For Tab = 1000, Ted = 500, e = 0.50. Without the regen¬ 
erator, the efficiency would have been 


e 


W g b — W dc 

H da + H a b 


RTa logeF- RT * log* 17 
778 l(T a - T d ) + RT a log. J- 6 

__ {Tg — Td) loge j 

T d )+ T a log,,/’ 


in which J = Vb/V a and (since ab and dc are isodiabatics) also 






Air Engines, Air Compression, Air Refrigeration 75 

= VjVd • For the temperatures above assumed, this becomes 

1150 log J log J 

6 = 1232 + 2300 log ./ = 1.07 + 2 log J ' 

Then e increases as J increases. 

The mean effective pressure, with or without regenerator, is 

Wab — W dc R log e J(T a — Td) 

Vm - 144(Fc - Vi) - 14AV d (J - 1) • 

For the assumed conditions, this is 

53.36 X 2.3 X 500 log J 426 log J 
Pm ~ 14AVd(J - 1) “ V d (J - 1) • 

This decreases (for a fixed value of Vd) as J increases. Hence 
low mean effective pressures accompany high efficiencies. 

Hot-air engines with regenerators have given up to 25 per 
cent, efficiency even in small sizes, and have used as little as 1.7 
•lb. of coal per Ihp. per hr., but the mean effective pressures are 
low and the heat-transmitting surface at the furnace burns out 
rapidly. 

Prob. 119. In Art. 62, find efficiencies without regenerator 
for J = 4, 5, 10, 20, ».• (Ans., 0.27, 0.28, 0.33, 0.36, 0.50.) 
Prob. 120. Find Vdp m in the above. 

(Ans., 86, 75, 47, 28, indeterminate.) 








CHAPTER IV 


STEAM AND OTHER VAPORS 
Vapor Properties and Tables 

63. Heating of Liquid. Consider the fundamental equation, 
H — T + I + W, Art. 2, Equation (3). For liquids and their 
vapors, I (the disgregation work) is not equal to zero. The 
equation is usually written for such fluids, H = E + W, where 
E = T + I and comprises all effects of heat other than external 
work, and may be called the internal work, or gain of internal 

ENERGY. 

When 1 lb. of liquid at 32° F. is heated, a rise of temperature 
and (in general) an increase of volume occur. W is therefore 
positive. Its value is very small. There are few cases, in the 
instance of water, for example, where it exceeds 1 B.t.u. Hence 
in the heating of water E is nearly equal to the whole amount 
of heat supplied (see Art. 77). 

The heating of the liquid may be continued until it begins to 
boil. At atmospheric pressure, each liquid has its own definite 
boiling point. At a lower pressure, it will boil at a lower tempera¬ 
ture. At a higher pressure, it will boil at a higher temperature. 
Fig. 20 shows the temperatures at which boiling occurs, for 
various liquids, at various pressures. Water appears here 
(with one exception) as having the highest boiling temperatures, 
but there are many liquids less volatile than water. All curves 
show that as the temperature increases, the pressure increases 
more and more rapidly. Therefore if high temperatures are 
desired in a steam engine (so as to attain high Carnot efficiency) 
high pressures almost necessarily result. There is an exception 
to this, however: see Art. 83. 

The HEAT OF THE LIQUID is THE QUANTITY OF HEAT NECES¬ 
SARY TO RAISE THE TEMPERATURE OF 1 LB. OF LIQUID FROM 

76 


Steam and Other Vapors 


77 


32° F. to the boiling temperature. Its symbol is h. The 
boiling temperature depends upon the pressure at which boiling 
occurs. There is a definite temperature of boiling for every 



pressure (Fig. 20) of a given liquid.* The temperature 32° F. 
is arbitrarily taken as a starting point. The value of h for all 

* As tabulated on pages 82, 83, values of h consider the pressure at the 
start to be that pressure at which the boiling point is 32°, viz., 0.0886 lb. per 
sq. in. The pressure then increases as the water is heated, until the boiling 
temperature is reached. 






























































































































78 


Thermodynamics, Abridged 


liquids is at this temperature, consequently, zero. For every 
different pressure at which boiling occurs, there will be a different 
value of h. 

The boiling point of ammonia at atmospheric pressure is 
— 27° F. This is below 32° F., hence the value of h at this 
temperature is negative. If t = temperature at which boiling 
occurs, s = mean specific heat of liquid from 32° to t°, h = 
s(t — 32). The value of s is generally somewhat variable. 

Prob. 121. Compute h for water when t = 212°. 

(Ans., 180 B.t.u.) 

Prob. 122. When water boils under 100 lb. absolute pressure, 
its temperature is 327.8° and h = 298.3. Find s from 32° to 
327.8°. (Ans., 1.008.) 

Prob. 123. The mean specific heat of liquid ammonia from 
5° to 32° is 1.052. Find h at 5°. (Ans., — 28.4 B.t.u.) 

64. Vaporization. If more heat is added after the liquid 
reaches its boiling temperature, vapor is formed. No vapor 
can be formed until the liquid has reached boiling temperature. 
During the formation of the vapor, it and the liquid are at the 
same temperature. This temperature is that at which the liquid 
boils at the existing pressure. The temperature of the vapor 
cannot be increased above this as long as any liquid is present. 
The removal of heat from a vapor (if kept at constant pressure) 
does not lower its temperature, but liquefies it, or part of it. 

A pound of liquid completely evaporated makes a pound of 
vapor. The vapor has a definite specific volume, v. Its 
temperature we have shown to be t. The whole operation of 
converting the pound of already boiling liquid into vapor con¬ 
sumes or requires the heat L, defined as the latent heat of 
vaporization, all of which is supplied at constant tempera¬ 
ture, t, of the fluid. 

The kind of vapor thus formed is called saturated vapor 
or dry vapor. At a given pressure, saturated vapor has the 
lowest possible temperature and specific volume of any vapor. 
Wet vapor may be regarded as incompletely evaporated liquid. 


Steam and Other Vapors 


79 


That is, for example, from 1 lb. of liquid at the boiling point, 
| lb. of saturated vapor may have been formed, so that the 
resulting fluid is a mixture of equal parts of vapor and boiling 
liquid. The temperature and specific volume of the vapor in 
this mixture will be t and v. Superheated vapor is formed by 
adding still more heat to saturated vapor. This can only be 
done after all of the liquid is evaporated. The temperature, t' y 
of superheated vapor is always higher than t: its volume, v', is 
greater than v. It behaves like an imperfect gas. 

Prob. 124. A saturated vapor exists at the pressure p = 100 
and the temperature t = 327.8. The pressure is increased 
while the temperature remains constant. What happens? 
(Ans., the vapor must liquefy.) If the pressure is lowered 
while the temperature remains constant, what happens? (Ans., 
the vapor becomes superheated.) 

65. Variation of L. In general, L decreases at high pressures. 
This does not mean that it is easier to generate vapor at high 
pressure, because as the pressure increases, the boiling tempera¬ 
ture increases and hence h increases. The increase in h is more 
than the decrease in L, in the case of steam. The total heat 
necessary to generate dry saturated vapor from 1 lb. of liquid 
at 32° is called the total heat, H, and is equal to h + L. This 
quantity increases, for steam, as the pressure increases. It 
always requires more heat to transform water (at a given initial 
temperature) to dry steam, when the pressure is high. The 
amount of heat required to vaporize liquid already at the 
boiling point decreases with increase of pressure. This quan¬ 
tity we call L. 

At moderate pressures and for most vapors, L is greater than h. 
Thus for water boiling at 212° F., L = 970.4 while h = 180. 
As the pressure increases, h increases and L decreases. Finally a 
pressure is reached at which L = 0. The temperature at which 
boiling then occurs is called the critical temperature. This, 
for steam, is 689° F. At the critical temperature, the liquid 
(brought to the boiling point) passes to the condition of saturated 


80 


Thermodynamics, Abridged 


vapor without any further addition of heat. The liquid and 
vapor states thus merge into each other. If a liquid is heated 
above its critical temperature it becomes a gas. It cannot 
again be liquefied by mere increase of pressure until its tempera¬ 
ture has first been reduced below the critical point. This is 
of importance in connection with carbon dioxide, which has a 
critical temperature around 88° F. This substance is used as a 
fluid in refrigerating plants on some vessels. The refrigerating 
fluid must be condensed in one part of the operation. If CO2 
is used, its temperature must then be brought below 88°. 
Conditions may arise in tropical waters where this would be- 
impossible. The C0 2 refrigerating machine cannot then be 
used. 

In the equation, heat = E + W (Art. 63), as applied to 
vaporization alone, how much of the heat of vaporization, L, 
is an E effect, and how much a W effect? We are considering 
vaporization alone: the liquid is already at the boiling point: 
L = E + W. Of the heat added during vaporization, W units 
are used in performing external work, and E units in performing 
internal work, or increasing the internal energy. Take the case 
of steam formed from water at 400°, and therefore (by the 
steam table) at 247.1 lb. pressure. The specific volume of the 
water at 400° is 0.0187. During vaporization, this is increased 
to v = 1.872. The external work part of L is therefore W L 
= pressure X increase of volume = 

144 X 247.1(1.872 - 0.0187) D . n _ 

1 - 71 - 7 Q — 84.2 13.t.u. 

1 1 o 

The E effect must be L — 84.2. The value of L given by the 
steam table is 827.2. Hence the E effect is 743: much larger 
than the W effect. The symbol r is used for the E effect in¬ 
cluded in L. Since h is nearly all E effect (Art. 63), H = h + L 
consists approximately of an E effect equal* to h + r and a W 
effect nearly equal to 144 — v w ) -f- 778, where v w = specific 
volume of liquid at the boiling point. 

* The exact value is given in Art. 77. 



Steam and Other Vapors 


81 


Prob. 125. Using the steam table, pages 82, 83, take values 
of p, v, L, as given at 327.8° F., and check the value given for r. 
Note v w = 0.01774 from Fig. 22 A. 

66 . The Steam Table. The steam table, given on the follow¬ 
ing pages, is a typical vapor-property table. Similar tables exist 
for other vapors like ammonia, S0 2 , C0 2 : but the properties of 
many vapors are only imperfectly established. The basis or 
“argument” of the table is either the pressure or the tempera¬ 
ture. For a given value of the pressure, p, t is the Fahrenheit 
temperature at which boiling occurs. The specific volume of 
the vapor is v, which is less for high than for low pressures. The 
heat quantities are for 1 lb. of fluid, originally liquid at 32° and 
at 0.0886 lb. pressure. The meaning of h, L, H and r has been 
explained. All tables have an elaborate experimental basis, 
and for saturated steam, the differences between the various 
existing tables are of small importance. The value of H here 
used is based on the Davis formula, 

H = 1150.3 + 0.37450 - 212) - 0.000550 - 212) 2 . (38) 

Prob. 126. From the Davis formula, find H when t = 212°. 

(Ans., 1150.3.) 

Prob. 127. Compute H when t = 312°. (Ans., 1182.25.) 

Prob. 128. From the table, what is the mean specific heat of 
water between 32° and 312°? (Ans., 1.007.) 

Prob. 129. What is the approximate value of E in the forma¬ 
tion of 1 lb. of steam at 80 lb. pressure from water at 32° F.? 
(Ans., 1101.8.) What is the value of W during vaporization? 
(Ans., 80.5 B.t.u.) 

Entropy 

67. A New Diagram. Art. 33 gave a graphical method of 
representing heat absorbed, of which we have not made much 
use. The subtended area under a path on the PV diagram, 
representing external work done, has on the other hand been 
used repeatedly. The magnitude of this area for specific pro¬ 
cesses is easily computed, because three of its boundaries are 


0 . 

0 , 

0 , 

0 . 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

14. 

15 

16 

17 

18 

19 

20 

21 

22 

23 

24 

25 

26 

27 

28 

29 

30 

31 

32 

33 

34 

35 

36 

37 

38 

39 

40 

41 

42 

43 

44 

45 

46 

47 

48 

49 

50 


Properties of Saturated Steam 

used from Steam Tables and Diagrams, by Marks and Davis, with 
e permission of the publishers, Messrs. Longmans, Green, & Co.) 


t 


32 

70 

80 

90 


101.83 

126.15 

141.52 

153.01 

162.28 


170.06 

176.85 

182.86 
188.27 
193.22 


197.75 

201.96 

205.87 

209.55 


.0 

216.3 

219.4 

222.4 

225.2 
228.0 
230.6 

233.1 

235.5 
237.8 

240.1 

242.2 

244.4 

246.4 

248.4 

250.3 


252.2 
254.1 
255.8 
257.6 

259.3 


161.0 

:62.6 

:64.2 

165.8 

67.3 


268.7 
270.2 

271.7 
273.1 
274.5 


275.8 
277.2 
278.5 

279.8 
281.0 


V 

h 

L 

II 

3294 

0 

1073.4 

: 1073.4 

871 

38.06 

1052.3 

; 1090.3 

636.8 

48.0 

1046.7 

1094.8 

469.3 

58.0 

1041.2 

! 1099.2 

333.0 

69.8 

1034.6 

1104.4 

173.5 

94.0 

1021.0 

1115.0 

118.5 

109.4 

1012.3 

1121.6 

90.5 

120.9 

1005.7 

1126.5 

73.33 

130.1 

1000.3 

1130.5 

61.89 

137.9 

995.8 

1133.7 

53.56 

144.7 

991.8 

1136.5 

47.27 

150.8 

988.2 

1139.0 

42.36 

156.2 

985.0 

1141.1 

38.38 

161.1 

982.0 

1143.1 

35.10 

165.7 

979.2 

1144.9 

32.36 

169.9 

976.6 

1146.5 

30.03 

173.8 

974.2 

1148.0 

28.02 

177.5 

971.9 

1149.4 

26.79 

180.0 

970.4 

1150.4 

26.27 

181.0 

969.7 

1150.7 

24.79 

184.4 

967.6 

1152.0 

23.38 

187.5 

965.6 

1153.1 

22.16 

190.5 

963.7 

1154.2 

21.07 

193.4 

961.8 

1155.2 

20.08 

196.1 

960.0 

1156.2 

19.18 

198.8 

958.3 

1157.1 

18.37 

201.3 

956.7 

1158.0 

17.62 

203.8 

955.1 

1158.8 

16.93 

206.1 

953.5 

1159.6 

16.30 

208.4 

952.0 

1160.4 

15.72 

210.6 

950.6 

1161.2 

15.18 

212.7 

949.2 

1161.9 

14.67 

214.8 

947.8 

1162.6 

14.19 

216.8 

946.4 

1163.2 

13.74 

218.8 

945.1 

1163.9 

13.32 

220.7 

943.8 

1164.5 

12.93 

222.6 

942.5 

1165.1 

12.57 

224.4 

941.3 

1165.7 

12.22 

226.2 

940.1 

1166.3 

11.89 

227.9 

938.9 

1166.8 

11.58 

229.6 

937.7 

1167.3 

11.29 

231.3 

936.6 

1167.8 

11.01 

232.9 

935.5 

1168.4 

10.74 

234.5 

934.4 

1168.9 

10.49 

236.1 

933.3 

1169.4 

10.25 

237.6 

932.2 

1169.8 

10.02 

239.1 

931.2 

1170.3 

9.80 

240.5 

930.2 

1170.7 

9.59 

242.0 

929.2 

1171.2 

9.39 

243.4 

928.2 

1171.6 

9.20 

244.8 

927.2 

1172.0 

9.02 

246.1 

926.3 

1172.4 

8.84 

247.5 

925.3 

1172.8 

8.67 

248.8 

924.411173.2 

8.51 

250.1 

923.511173.6 


n w ■ n e I n. 


1019.3 

0 

2.1832 

2.1832 

994.0 

0.0745 

1.9868 

2.0613 

987.4 

0.0932 

1.9398 

2.0330 

980.8 

0.1114 

1.8944 

2.0058 

972.9 

0.1327 

1.8427 

1.9754 

956.7 

0.1749 

1.7431 

1.9180 

946.4 

0.2008 

1.6840 

1.8848 

938.6 

0.2198 

1.6416 

1.8614 

932.4 

0.2348 

1.6084 

1.8432 

927.0 

0.2471 

1.5814 

1.8285 

922.4 

0.2579 

1.5582 

1.8161 

918.2 

0.2673 

1.5380 

1.8053 

914.4 

0.2756 

1.5202 

1.7958 

910.9 

0.2832 

1.5042 

1.7874 

907.8 

0.2902 

1.4895 

1.7797 

904.8 

0.2967 

1.4760 

1.7727 

902.0 

0.3025 

1.4639 

1.7664 

899.3 

0.3081 

1.4523 

1.7604 

897.6 

0.3118 

1.4447 

1.7565 

896.8 

0.3133 

1.4416 

1.7549 

894.4 

0.3183 

1.4311 

1.7494 

892.1 

0.3229 

1.4215 

1.7444 

889.9 

0.3273 

1.4127 

1.7400 

887.8 

0.3315 

1.4045 

1.7360 

885.8 

0.3355 

1.3965 

1.7320 

883.9 

0.3393 

•1.3887 

1.7280 

882.0 

0.3430 

1.3811 

1.7241 

880.2 

0.3465 

1.3739 

1.7204 

878.5 

0.3499 

1.3670 

1.7169 

876.8 

0.3532 

1.3604 

1.7136 

875.1 

0.3564 

1.3542 

1.7106 

873.5 

0.3594 

1.3483 

1.7077 

872.0 

0.3623 

1.3425 

1.7048 

870.5 

0.3652 

1.3367 

1.7019 

869.0 

0.3680 

1.3311 

1.6991 

867.6 

0.3707 

1.3257 

1.6964 

866.2 

0.3733 

1.3205 

1.6938 

864.8 

0.3759 

1.3155 

1.6914 

863.4 

0.3784 

1.3107 

1.6891 

862.1 

0.3808 

1.3060 

1.6868 

860.8 

0.3832 

1.3014 

1.6846 

859.5 

0.3855 

1.2969 

1.6824 

858.3 

0.3877 

1.2925 

1.6802 

857.1 

0.3899 

1.2882 

1.6781 

855.9 

0.3920 

1.2841 

1.6761 

854.7 

0.3941 

1.2800 

1.6741 

853.6 

0.3962 

1.2759 

1.6721 

852.4 

0.3982 

1.2720 

1.6702 

851.3 

0.4002 

1.2681 

1.6683 

850.3 

0.4021 

1.2644 

1.6665 

849.2 

0.4040 

1.2607 

1.6647 

848.1 

0.4059 

1.2571 

1.6630 

847.1 

0.4077 

1.2536 

1.6613 

846.1 

0.4095 

1.2502 

1.6597 

845.0 

0.4113 

1.2468 

1.6581 




























Stea^ and Other Vapors 


83 


Properties of Saturated Steam—Continued 

(Condensed from Steam Tables and Diagrams, by Marks and Davis, with 
the permission of the publishers, Messrs. Longmans, Green, & Co.) 


p 

t 

V 

h 

L 

H 

r 

'R'W 

n e 

7l a 

51 

282.3 

8.35 

251.4 

922.6 

1174.0 

844.0 

0.4130 

1.2435 

1.6565 

52 

283.5 

8.20 

252.6 

921.7 

1174.3 

843.1 

0.4147 

1.2402 

1.6549 

53 

284.7 

8.05 

253.9 

920.8 

1174.7 

842.1 

0.4164 

1.2370 

1.6534 

54 

285.9 

7.91 

255.1 

919.9 

1175.0 

841.1 

0.4180 

1.2339 

1.6519 

55 

287.1 

7.78 

256.3 

919.0 

1175.4 

840.2 

0.4196 

1.2309 

1.6505 

56 

288.2 

7.65 

257.5 

918.2 

1175.7 

839.3 

0.4212 

1.2278 

1.6490 

57 

289.4 

7.52 

258.7 

917.4 

1176.0 

838.3 

0.4227 

1.2248 

1.6475 

58 

290.5 

7.40 

259.8 

916.5 

1176.4 

837.4 

0.4242 

1.2218 

1.6460 

59 

291.6 

7.28 

261.0 

915.7 

1176.7 

836.5 

0.4257 

1.2189 

1.6446 

60 

292.7 

7.17 

262.1 

914.9 

1177.0 

835.6 

0.4272 

1.2160 

1.6432 

61 

293.8 

7.06 

263.2 

914.1 

1177.3 

834.8 

0.4287 

1.2132 

1.6419 

62 

294.9 

6.95 

264.3 

913.3 

1177.6 

833.9 

0.4302 

1.2104 

1.6406 

63 

295.9 

6.85 

265.4 

912.5 

1177.9 

833.1 

0.4316 

1.2077 

1.6393 

64 

297.0 

6.75 

266.4 

911.8 

1178.2 

832.2 

0.4330 

1.2050 

1.6380 

65 

298.0 

6.65 

267.5 

911.0 

1178.5 

831.4 

0.4344 

1.2034 

1.6368 

66 

299.0 

6.56 

268.5 

910.2 

1178.8 

830.5 

0.4358 

1.2007 

1.6355 

67 

300.0 

6.47 

269.6 

909.5 

1179.0 

829.7 

0.4371 

1.1972 

1.6343 

68 

301.0 

6.38 

270.6 

908.7 

1179.3 

828.9 

0.4385 

1.1946 

1.6331 

69 

302.0 

6.29 

271.6 

908.0 

1179.6 

828.1 

0.4398 

1.1921 

1.6319 

70 

302.9 

6.20 

272.6 

907.2 

1179.8 

827.3 

0.4411 

1.1896 

1.6307 

71 

303.9 

6.12 

273.6 

906.5 

1180.1 

826.5 

0.4424 

1.1872 

1.6296 

72 

304.8 

6.04 

274.5 

905.8 

1180.4 

825.8 

0.4437 

1.1848 

1.6285 

73 

305.8 

5.96 

275.5 

905.1 

1180.6 

825.0 

0.4449 

1.1825 

1.6274 

74 

306.7 

5.89 

276.5 

904.4 

1180.9 

824.2 

0.4462 

1.1801 

1.6263 

75 

307.6 

5.81 

277.4 

903.7 

1181.1 

823.5 

0.4474 

1.1778 

1.6252 

80 

312.0 

5.47 

282.0 

900.3 

1182.3 

819.8 

0.4535 

1.1665 

1.6200 

85 

316.3 

5.16 

286.3 

897.1 

1183.4 

816.3 

0.4590 

1.1561 

1.6151 

90 

320.3 

4.89 

290.5 

893.9 

1184.4 

813.0 

0.4644 

1.1461 

1.6105 

95 

324.1 

4.65 

294.5 

890.9 

1185.4 

809.7 

0.4694 

1.1367 

1.6061 

100 

327.8 

4.429 

298.3 

888.0 

1186.3 

806.6 

0.4743 

1.1277 

1.6020 

105 

331.4 

4.230 

302.0 

885.2 

1187.2 

803.6 

0.4789 

1.1191 

1.5980 

110 

334.8 

4.047 

305.5 

882.5 

1188.0 

800.7 

0.4834 

1.1108 

1.5942 

115 

338.1 

3.880 

309.0 

879.8 

1188.8 

797.9 

0.4877 

1.1030 

1.5907 

120 

341.3 

3.726 

312.3 

877.2 

1189.6 

795.2 

0.4919 

1.0954 

1.5873 

125 

344.4 

3.583 

315.5 

874.7 

1190.3 

792.6 

0.4959 

1.0880 

1.5839 

130 

347.4 

3.452 

318.6 

872.3 

1191.0 

790.0 

0.4998 

1.0809 

1.5807 

140 

353.1 

3.219 

324.6 

867.6 

1192.2 

785.0 

0.5072 

1.0675 

1.5747 

150 

358.5 

3.012 

330.2 

863.2 

1193.4 

780.4 

0.5142 

1.0550 

1.5692 

160 

363.6 

2.834 

335.6 

858.8 

1194.5 

775.8 

0.5208 

1.0431 

1.5639 

170 

368.5 

2.675 

340.7 

854.7 

1195.4 

771.5 

0.5269 

1.0321 

1.5590 

180 

373.1 

2.533 

345.6 

850.8 

1196.4 

767.4 

0.5328 

1.0215 

1.5543 

190 

377.6 

2.406 

350.4 

846.9 

1197.3 

763.4 

0.5384 

1.0114 

1.5498 

200 

381.9 

2.290 

354.9 

843.2 

1198.1 

759.5 

0.5437 

1.0019 

1.5456 

210 

386.0 

2.187 

359.2 

839.6 

1198.8 

755.8 

0.5488 

0.9928 

1.5416 

220 

389.9 

2.091 

363.4 

836.2 

1199.6 

752.3 

0.5538 

0.9841 

1.5379 

230 

393.8 

2.004 

367.5 

832.8 

1200.2 

748.8 

0.5586 

0.9758 

1.5344 

240 

397.4 

1.924 

371.4 

829.5 

1200.9 

745.4 

0.5633 

0.9676 

1.5309 

250 

401.1 

1.850 

375.2 

826.3 

1201.5 

742.0 

0.5676 

0.9600 

1.5276 

2947 

689.0 

0.05 

? 

0 

? 

0 

? 

0 

? 
























84 


Thermodynamics, Abridged 


straight lines. We are now to develop a similarly simple area 
for the representation of quantities of heat absorbed or emitted. 
The new diagram may be conceived as having been produced 
by a mechanism “ interlocked ” with the indicator, so that while 



the pencil of the latter is describing lines the areas under which 
represent external work, the imaginary pencil of the new diagram 
will simultaneously sketch corresponding lines, areas under which 
will represent heat. 































































































Steam and Other Vapors 


85 


The ordinate of the diagram is to be absolute temperature. 
In Fig. 21, then, T is the ordinate and abed represents the heat 
absorbed or emitted along ab , or briefly abed = H a b- If nb is 
described from left to right, H a b is +: if from right to left, H a b 
is —. Bearing in mind that subtended areas represent 
HEAT and ORDINATES ARE ABSOLUTE TEMPERATURES, answer the 
following: 

Prob. 130. What is the shape of an isothermal on this 
diagram? 

Prob. 131. What is the only possible shape of an adiabatic? 
Why? 

Prob. 132. What is the shape of a Carnot cycle? Sketch 
and letter the areas representing Hjk and Hi m . Which is plus 
and which minus if the cycle is clockwise? What area represents 
the net external work done in the Carnot cycle? (ZW = ZH 
= heat received — heat rejected.) Efficiency is ZH divided by 
all of the + heat. In this case, it is the quotient of what two 
areas? Of what two lengths (in rectangles of equal width, areas 
are proportional to lengths). State the efficiency in terms of 
T jk and T m i and prove the statement. 

68 . The Abscissa of the Diagram. Consider the heating of 1 
lb. of water from 32° F. or 492° absolute. Heat it 1 °. The 
heat absorbed is 1.01 B.t.u. (the specific heat of water is not 
exactly 1.0). Let o, Fig. 21 , represent the condition of the 
water before heating. A path representing the heating process 
must be subtended by an area of 1.01 B.t.u. If we assume the 
path to be a straight line, the subtended area is a trapezoid of 
area 1.01 and parallel vertical sides of lengths 492 and 493 
(absolute temperatures). Its horizontal width is therefore 
area -f- mean height = 1.01 7 - 492.5 = 0.00205. The process 
is represented by the line op: the coordinates of 0 being 492, 0 
(the latter fixed as an arbitrary starting point) and those of p 
being 493, 0.00205. 

Heat this water another degree. The heat supplied is this 
time 1.0 B.t.u., the mean temperature 493.5, the horizontal 


86 


Thermodynamics, Abridged 


width of the trapezoid 0.00203, and the coordinates of q are 494 
and 0.00205 + 0.00203 = 0.00408. 


Prob. 133. In heating from 34° to 35° F., and also from 35° 
to 36° F., 1.01 B.t.u. are supplied. Locate points r and s, Fig. 
21. (Ans., abscissas are 0.00613 and 0.00817.) 


The whole operation from 32° to 36° is represented by os, 
which is nearly a straight line. This method is approximate, 
involving the assumption that each of the segments op, pq, qr, 
rs, is a straight line. The error due to this is exceedingly small. 

Consider the process tu. The area under this curve repre¬ 
sents H tu . Consider a very small portion of the curve, along 
which the temperature may be assumed to be constant. Call 
this temperature T. The narrow area under the small portion 
of the curve is dH and the width of this area is dH/T. The 
horizontal component or abscissa of the whole path tu is then 


71 in — 7lu 711 — 



between proper limits. 


The dimension n is called the entropy. The only definition 
of entropy is J'(dH/T). It is the abscissa of our diagram, 
which is accordingly marked n. There is no necessary origin 
for entropy. The line os puts it at 32° F. or 492° absolute, but 
this need not always be the case. Only changes of entropy are 
important. The symbol n tu or the quantity n u — n t both mean 

the CHANGE OF ENTROPY Or the HORIZONTAL COMPONENT OF THE 

path from t to u. We must now establish specific values for 
J*(dHlT) for various processes. 

69. Change of Entropy at Constant or Definite Specific Heat. 
For the path tu, n u — n t — J'(dH/T). Suppose this path to be 
one along which the specific heat is constant = s. Then H 
— s(T 2 — Ti) for 1 lb. weight heated from Ti to T 2 or dH — sdT. 
Then 

X u sdT T 

~jT = s(log e T u — log e Tt) = S loge yT. (39) 


Note that T u and T t are absolute temperatures. Suppose the 
specific heat to be variable: for example, Si = a + b T. Then 


Steam and Other Vapors 


87 


r u adT C u T 

~jT + J bdT = a loge ^r+ b(T u - 7 1 *)- 

The case of water comes in neither of these classes. Its specific 
heat is too irregularly variable to be expressed as a linear function 
of the temperature. An approximation can be made, for water, 
by taking s as the mean value over the temperature range 
specified, in Equation (39). 

Thus for water heated from 32° to 36° F. the heat expenditure 
is 4.03 B.t.u. Take s, then, at 4.03 -f- 4 = 1.0075. Then by 
Equation (39), 

496 

n 36 - n Z2 = 1.0075 X 2.3026* log ^ = 0.0082. 

This result is comparable with that reached in Prob. 133 and 
used in plotting the point s, Fig. 21. It should be noted that 
with a positive specific heat the path must slope upward to the 
right and downward to the left: that low values of the specific 
heat lead to steep paths: and that any path of constant specific 
heat becomes steeper as it ascends. 

Prob. 134. Given T u =2T t . Find n tu when s = 0.1689, 
0.2375, 0.5, 1.0. (Ans., 0.117, 0.164, 0.347, 0.693.) 

Prob. 135. What is the gain of entropy in heating 1 lb. of 
water from 32° F. to 312° F.? Note: take s = h -r- (t — 32). 

(Ans., 0.4535.) 

70. Entropy at Various Specific Heats. For an adiabatic, 
5=0 and the path is vertical. The table in Art. 29 mentioned 
the adiabatic as a path of constant entropy. For an isothermal, 
s = oo and the path is horizontal. For all constant values of s 
between 0 and + oo, the curve is a logarithmic curve. Sev¬ 
eral paths are plotted in the small diagram in the upper right- 
hand corner of Fig. 21. Equation (39) may be used to determine 
either end-points or intermediate points of these paths. 

Prob. 136. If 5 = 1, T t = 500, T u = 750, find n tu . 

(Ans., 0.406.) 

* The use of the modulus permits employing a table of common logarithms 
instead of Napierian logarithms. 

7 


88 


Thermodynamics, Abridged 


71. Entropies of Vapor. The entropy of the liquid, n w , is 

the CHANGE OF ENTROPY DUE TO THE HEATING OF 1 LB. OF LIQUID 

from 32° F. to the boiling point. It was computed approx¬ 
imately for water with t = 312° F. in Prob. 135. Similar com¬ 
putations taken for successive short ranges of temperature give 
all of the values of n w which appear in the steam table, third 
column from the last. 

The ENTROPY OF VAPORIZATION, W e , is the CHANGE OF ENTROPY 
of 1 lb. of fluid which occurs during vaporization. Vaporiza¬ 
tion takes place at constant temperature. A constant tempera¬ 
ture process, in Fig. 21, is represented by a horizontal line. The 
subtended area is a rectangle. The width of the rectangle is 
the change of entropy. It is equal to the area divided by the 
height or to the heat absorbed divided by the absolute tempera¬ 
ture. Hence n e = L -f- (£+ 460). Values are given in the 
column of the steam table next to the last. 

The TOTAL ENTROPY, OR ENTROPY OF 1 LB. OF VAPOR, is U s 
= n w + n e . Values appear in the last column of the table for 
steam. 

Prob. 137. Compute n e and n s for steam at 312° F. (See 
Prob. 135. Ans., n e = 1.1665, n s = 1.62.) 

72. Entropy Diagram for Steam. In Fig. 22, the ordinate 
scale is converted into Fahrenheit temperature. As in Fig. 21, 
the lower part of the diagram is omitted. All heat areas are 
to be understood as carried down to — 460° F. Water is to be 
heated from 32°. The state is a, taken on the vertical axis. 
The path ac represents the heating of water from 32° F. to 
150° F. The abscissa of this path (length ot) is (n w ) i 50 = 0.2149. 
The water is then further heated to 327.8° F. along cb. The 
abscissa of the whole path ab is os = (n w ) 32 7.8 = 0.4743. The 
gain of entropy in heating water from 150° to 327.8° is os — ot 
= ts = 0.4743 — 0.2149 = 0.2594. The whole curve ab may 
thus be plotted from the steam table as accurately and as com¬ 
pletely as may be desired. 


Steam and Other Vapors 


89 



Prob. 138. What is the abscissa of that part of the curve ab 
which terminates at 212° F.? (Ans., 0.3118.) How much 
entropy is gained when 1 lb. of w^ater is heated from 212° to 
312°? (Ans., 0.1417.) 

Suppose the pressure to be such that the water begins to boil 
at the point b, Fig. 22. Here the temperature is 327.8° F. 
Vaporization takes place at constant temperature, and is there¬ 
fore represented by the horizontal line from b. At some point d, 
on this line, vaporization will be completed. The length of the 
line bd represents the change of entropy during vaporization, or 
briefly the entropy of vaporization. The symbol for this 
quantity is n e . Its value for the assumed condition is 

(^e) 327.8 = bd = (L/T)z27.8 = 1.1277. 

If the pressure had been different, so that vaporization had 
taken place at 150° F., the path representing vaporization would 
have been ce. The dotted line de joining the end-points of 
vaporization paths like ce and bd is a locus of completely vapor- 




























































































































































90 


Thermodynamics, Abridged 


ized states. It represents every possible kind of dry steam. Its 
abscissa at any point is n s . The curves ab and ed meet at the 
CRITICAL TEMPERATURE. 

All heat quantities are measured above 32° F. as a starting 
point. The area under ac ( acgf , carried down to the — 460° F. 
level) therefore represents the heat of the liquid at c, abbrevi¬ 
ated h e . The area under ce ( cejg ) similarly carried down, repre¬ 
sents the heat consumed in converting liquid at c into dry vapor 
at e, abbreviated as L ce . The sum of these two areas represents 
H e , the total heat of dry steam at e, above 32° F. 

Prob. 139. Locate d, where t = 327.8. 

(Ans., T = 787.8, n = (n s ) 327>8 = 1.602.) 

Any point on ab represents boiling water. Any point on de 
represents completely evaporated (“ dry ”) steam. Any point 
between these two curves represents a mixture of steam and 
boiling water, i.e., wet steam. If the point is close to de, the 
mixture is nearly dry steam. If the point is close to ab, the 
mixture is mostly water. If the state moves from a point near 
ab toward de, water is being evaporated. If the movement is in 
the reverse direction, steam is being condensed. 

An adiabatic process is represented by a vertical line. If hot 
water (as at b) expands to a lower pressure, adiabatically (path 
bn), some of it will be vaporized, because the point n is farther 
from ab than is the point b. This actually occurs when the blow-¬ 
off cock of a boiler is opened. If dry steam is expanded adiabat¬ 
ically (path dp), some of it will condense. If very wet steam is 
compressed adiabatically (path nb) it will be partially or wholly 
liquefied. If nearly dry steam is sufficiently compressed adia¬ 
batically (path pd) it will be made wholly dry. A point m, 
half way between ab and de, represents steam “ 50 per cent, 
dry,” i.e., it represents a mixture in which half the water has 
been evaporated while the other half is still liquid. 

Prob. 140. Locate the point u, Fig. 22, representing steam 
50 per cent, dry at 150° F. (Ans., entropy at the point c = 
0.2149, entropy at the point e = 1.8674, n u = 0.2149 + §(1.8674 
- 0.2149) = 1.0412.) 


Steam and Other Vapors 


91 


The “ constant dryness ” curves mu, vw may be plotted by 
computations made as in Prob. 140, or may be obtained graph¬ 
ically by properly dividing horizontal distances between ab 
and de. Thus bm = md, mv = vd, cu = ue, uw = we. 

Suppose heat to be added when the state is d. There being 
no liquid present, the temperature will rise; i.e., the path from d 
will be upward. Since heat is supplied, there must be a sub¬ 
tended area. Hence the path will not be vertical. Since the 
sign of the heat supplied is +, the path must be from left to 
right: i.e., it will slope, so to speak, toward the northeast. If 
the specific heat during this operation were constant, the path 
would be a logarithmic curve, indicated by Equation (39), like dx. 
The operation dx is one of superheating. Superheating implies 
the addition of heat (no liquid being present), increase of temper¬ 
ature and increase of entropy. All points lying to the right of de 
represent superheated vapor. If superheated steam expands 
adiabatically (path xy ) it becomes less superheated, then dry and 
finally wet. If dry steam (or nearly dry steam) is compressed 
adiabatically (path er) it is superheated. The heat absorbed 
during the superheating operation dx is represented by the area 
under dx, carried down to the — 460° F. level. 

73. Factor of Evaporation. In usual practice, we do not start 
the formation of steam with water at 32°, but at some higher 
temperature, to. The heat expenditure for warming from this 
temperature to t\ is hi — ho B.t.u. The total expenditure of 
heat in forming dry steam at ti° is then hi — h 0 + Li = Hi — ho 
B.t.u. This varies with the pressure, pi, at which evaporation 
occurs, and with the temperature t 0 at which water is supplied 
in the first place. This temperature is called the feed-water 
temperature. The generation of 1 lb. of dry steam from feed 
water therefore represents a variable quantity of heat. 

For the purpose of comparison, the standard condition to 
which all others are referred is thus defined: the feed temperature 
is 212° F. Hence ho = 180. Also, the steam pressure is normal 
atmospheric pressure. Hence hi is also 180. Then the total 


92 


Thermodynamics, Abridged 


heat expenditure per lb. of steam formed is hi — h 0 + L\ = L^n 
= 970.4. This condition is described as from (a feed water 
temperature of) and at (a steam pressure corresponding with) 
212 ° F. In Fig. 22, the area under zA represents L 2 u, the heat 
expenditure necessary to generate 1 lb. of steam “from and at 
212 ° F.” The area under chd represents the heat expenditure 
when the feed temperature is t c and the steam pressure is the 
pressure which corresponds with the temperature The 

quotient of the latter area by the former is called the factor of 
evaporation, which has the value 

h- K + ha H d - K 

970.4 970.4 

The factor of evaporation is the ratio of the heat actually em¬ 
ployed to make steam, divided by the heat necessary from and 
at 212° F. A high value of F means costly steam. 


Some Equivalents 


Mechanical Units 

1 ft. lb. 

778 ft. lb. 

754 971 ft. lb. 


Heat Units 

= 1/778 B.t.u. 

= 1 B.t.u. 

= 1 lb. of water evaporated from and 
at 212° F. = 970.4 B.t.u. 


33 000 ft. lb. per min. = 1 hp. = 42.42 B.t.u. per min. 
1 980 000 ft. lb. per hr. = 1 hp. = 2545 B.t.u. per hr. 


Prob. 141. How many B.t.u. are required to heat 10 lb. of 
water from 212° F. to 312° F.? (Ans., 1020 B.t.u.) 

Prob. 142. How many B.t.u. are required to generate 1 lb. 
of dry steam at 80 lb. pressure from feed water at 90° F.? (Ans., 
1124.3.) What is the factor of evaporation? (Ans., 1.16.) 

Prob. 143. Find F when t 0 = 80, pi = 200. (Ans., 1.18.) 

Prob. 144. A boiler evaporates 10 lb. of water qer lb. of 
coal under the conditions of Prob. 142. Another boiler, using 
the same coal, evaporates 9.9 lb. of water per lb. of coal under 
the conditions of Prob. 143. Which is the better boiler? How 
much better? (Ans., the second boiler is 1.1 per cent, better.) 



Steam and Other Vapors 


93 


Wet Vapor: Vapor Processes 

74. Properties of Wet Vapor. If 1 lb. of a liquid is brought 
to the boiling point t and the pressure p and then the portion x 
lb. of this liquid is evaporated, the properties are: 

h, B.t.u., heat of liquid; for all of the liquid was brought to the 
boiling point: 

n w , entropy of liquid; for the substance moved along ab, Fig. 
22 , in the usual way: 

t, the common Fahrenheit temperature of the liquid and the 
vapor: 

xL, the heat of vaporization; for only x lb. were vaporized: 

xn e , the entropy of vaporization; for xL/T = xn e : 

h + xL, the total heat, which is the sum of the heats of liquid 
and of vaporization: 

n w + xn e , the total entropy, which is the sum of the entropies 
of liquid and of vaporization. 

The expression for specific volume of wet vapor is less obvious. 
The volume of 1 lb. of the liquid at the boiling point is v w . In 
a mixture weighing 1 lb., (1 — x) lb. of liquid are present. 
There are also present a: lb. of dry vapor, the total volume of 
which is xv. Then the volume of 1 lb. of the mixture is (1 — x)v w 
+ xv = v w + x(v — v w ). This is very nearly equal to xv, and 
may be so taken unless special note is made to the contrary. 

The external work of vaporization was shown to be 

W = ^{v-v w ) B.t.u. 


when dry vapor was formed. The quantity in the parenthesis 
is the increase of volume from liquid to dry vapor. The 
corresponding increase for wet vapor is x{v — v w ). The value 
of W for vaporization to wet vapor is consequently 


144p 

778 


x(v — v w ) 


which is x times the value for dry vapor. The internal work 
(E effect) in the vaporization to dry vapor was r = L — W. 
For wet vapor it is xL — xW = x(L — W) = xr. 


94 


Thermodynamics, Abridged 


Recapitulation: 


Properties that are the same 
for wet vapor as for dry 


h, Tlwy Pi t , 


Properties multiplied by x 


dry vapor, L, n e , 
wet vapor, xL, xn ey 


h 

xr> 


Other properties 


dry vapor, v, H , n s , 

wet vapor, + x(v — v w ), h + xL, n w + xn e , 

(approximately xv). 


The quantity x is the dryness fraction or proportion of 
dryness. It is the weight of dry vapor in 1 lb. of the mixture 
of dry vapor and liquid. 


Prob. 145. From values given in the steam table at 100 lb. 
pressure, compute all of the corresponding properties for steam 
at this pressure if 0.90 dry. (Ans., p = 100, t = 327.8, h = 
298.3, n w = 0.4743, xL = 799.2, xn e = 1.0149, xr = 725.94, 
v w + — v w ) = 3.9876 if v w = 0.0177, h + xL = 1097.5, n w 

+ xn e = 1.4892.) 

The factor of evaporation for a wet vapor, following Equa¬ 
tion (40), is 


F w = 


xL + h — h 0 
970.4 ’ 


(41) 


where quantities without subscripts refer to the vapor condition 
and ho to the feed water condition. 

Prob. 146. What is the factor of evaporation for the steam 
in Prob. 145, if that steam was generated from feed water at 
90° F.? (Ans., 1.07.) 

75. Volumes on the Tn Diagram. Suppose it to be required 
that we plot in Fig. 22 the locus of states where the volume is 
25 cu. ft. Write 

25 = v w + x(v — v w ), or briefly, xv. 

The values of v w in the following are from Fig. 22A. Those 
of v, n w and n e are from the steam table: 



Steam and Other Vapors 


95 


For t =50 75 100 125 150 175 200 215 

v = 1702 743 350.8 178.4 96.9 55.7 33.6 25.35 

v w = 0.01602 0.01606 0.01603 0.01623 0.01634 0.01648 0.01663 0.01673 
25_ v * 

x = -* =0.015 0.034 0.071 0.14 0.258 0.45 0.743 0.987 

v — v w 


n e = 2.0865 1.9631 1.8505 1.7475 1.6525 1.5645 1.4824 1.4354 

xn e = 0.031 0.0661 0.132 0.245 0.426 0.702 1.103 1.42 

n w = 0.0361 0.0840 0.1295 0.1730 0.2149 0.255 0.2937 0.3163 

n w + xn e = 0.0671 0.1501 0.2615 0.4180 0.6409 0.957 1.3967 1.7363 


Values in the last line, plotted against those in the first line, 
give the constant volume curve of 25 cu. ft. in Fig. 22. Curves 



Specific Volume v w 

Fig. 22a. Specific Volume of Water. 

for less than 25 cu. ft. lie above this: curves for larger volumes 
lie below it. The maximum volume on the diagram is toward 
the “south-east” corner. 

Prob. 147. Find the entropy of steam at 50° F. when its 
volume is 100 cu. ft. (Ans., 0.158.) Find also at 75°, 100°, 125°. 
(Ans., 0.349, 0.658, 1.152.) 

76. Constant Heat Content. The total heat of wet steam is 
h + xL. This may have any value between rather wide limits, 
depending on p and x. Thus it may be 1000 B.t.u. when p = 100 
if x = (1000 - h)/L= 701.7/888 = 0.791. Then n w + xn e = 
0.4743 + (0.791 X 1.1277) = 1.366. From values found in this 

* Note that it is sufficiently accurate to take x = 25 / v . This- would not 
be a safe procedure, however, if we were plotting a line of very small volume. 







































96 Thermodynamics, Abridged 

way, the curve of constant heat content = 1000 B.t.u. is plotted 
in Fig. 22. 

Prob. 148. What are the dryness and entropy of steam to 
atmospheric pressure when the heat content is 1000 B.t.u.? 
(Ans., x = 0.845, n w + xn e = 1.53.) 

Prob. 149. If the steam in Prob. 148 emits 500 B.t.u. without 
change of pressure, what is then its dryness? (Ans., 0.33.) 
If it then emits 406 B.t.u. more, what does it become? (Ans., 
water at 126.15° F., if the pressure is allowed to fall after con¬ 
densation begins: within a small fraction of 1° of this tempera¬ 
ture if the pressure is kept at that of the atmosphere.) 

Prob. 150. A boiler generates from feed water at 90°, 20 000 
lb. of steam per hr. at 200 lb. pressure, 0.99 dry. The coal has 
a heat value of 11 316.7 B.t.u. per lb. and the efficiency of the 
boiler is 0.75. What weight of coal is burned per hr.? (Ans., 
2667 lb.) 

77. Work Done in Heating Liquid. The heat absorbed in the 
operation ab, Fig. 22, is hb per lb. of liquid. The external work 
done during this heating, and included in hb, is very small and 
may often be neglected. The method of computing it should 
be known. 

A pound of water has been heated from 32° F. and a pressure 
of 0.0886 lb. per sq. in. (Art. 63), until its pressure is that cor¬ 
responding with tb, or 100 lb. per sq. in. The external work is 
that due to this increase of pressure. It is equivalent to the 
work of lifting the pound of water to a height represented by 
the difference of the final and initial pressures. What would be 
this height? Suppose the water to be delivered into the base of a 
vertical tube of 1 sq. ft. area. If the height to the top of the 
tube (where the water flows off) is f ft., and the density of the 
water is d, the hydrostatic pressure at the base of the tube is 
simply the weight of water, or fd lb. per sq. ft. Or, inversely, 
the height in ft. is the pressure per sq. ft. divided by the density. 
At 100 lb. pressure per sq. in., the density of water is 56.4 lb. 


Steam and Other Vapors 


97 


per cu. ft. The external work of our process is equivalent to 
lifting 1 lb. of water to a height of (substantially) 100 X 144 
-T- 56.4 = 255 ft., or is 255 ft. lb. In general symbols, 

External Work = W h = B.t.u., (42) 

for v w is the reciprocal of the density. The gain of internal 
energy during the heating of the liquid is h — (pi^/5.403). The 
external work done in heating 1 lb. of liquid from c to b is (W h )b 
— (WT) C . Values of v w are given in Fig. 22 A. 

Prob. 151. Find W h at the critical temperature, where 
v w = 0.049. (Ans., 26.7 B.t.u.) 

78. Vaporization or Condensation. As shown in Art. 74, the 
heat absorbed in partial vaporization is xL B.t.u. per lb. For 
complete vaporization, put x = 1.0. Vaporization (partial) at 
constant pressure is represented by such a line as bv, Fig. 22. 
Condensation of steam initially at the point v would be repre¬ 
sented by the reverse line vb. The heat emitted is ( xL) v , and 
its sign is negative. Note that the isothermal and the constant 
pressure paths coincide when the fluid consists of a mixture of 
liquid and vapor. The external work done (Art. 74) is 144 
px(v — v w ) -r- 778 B.t.u. This is + for vaporization and x = 1 
if vaporization is complete. It is — for condensation. The 
internal work is + xr for vaporization and — xr for condensation. 

Prob. 152. How many B.t.u. are emitted when 1000 lb. of 
steam at 1 lb. pressure, 0.70 dry, are condensed and cooled to 
90°? (Ans., 736 020 B.t.u.) What is the exact loss of internal 
energy? Take v w = 0.01614 at 1 lb. and 0.0161 at 90°. (Ans., 
692 829 B.t.u.) 

79. Adiabatic Process. No heat is absorbed or emitted if the 
process is adiabatic. The external work is computed by writing 
as the law of a vapor adiabatic, initial entropy = final entropy: 
or, (Fig. 22) 

n B = n c , 

( n w + xn e ) b ~ (n w + xn e ) c- 


( 43 ) 




98 


Thermodynamics, Abridged 


This is one of the most useful equations of our subject. Either 
value of x may be 1.0. For an adiabatic, 

H = 0 = E+W, W = - E = E b - Eg, 
W= { h -5M +Xr ) B -( h -5M +Xr ) c - (44) 

This is very nearly equal to ( h + xr) B — (h + xr ) c, and in 
ordinary calculations should be so taken. Do not use this 
approximation, however, in Probs. 153-156. 

The work is + when the path is downward and — when it 
is UPWARD. 

On a pv diagram, the vapor adiabatic is a smooth curve which 
may be represented approximately by the equation PbVb° 
= pcVc a , but the value of a depends on the initial dryness: 
a — 1.035 + O.Izb. Then vc = VsipB/pc) 1 ' 0 and 

W = Pb ” b - f = 144 ( PB V B ~ g C - C V lb. 

a — l \ a— 1 ) 

Prob. 153. Steam at 250 lb. pressure, 0.99 dry, expands 
adiabatically to 1 lb. pressure. What is the dryness after 
expansion? (Ans., 0.7515.) What was the specific volume 
before expansion? (Ans., 1.832 if v w = 0.0187.) After expan¬ 
sion? (Ans., 250.) 

Prob. 154. In Prob. 153, find p^/5.403 at the initial condi¬ 
tion. (Ans., 0.87 B.t.u.) At the final condition, when v w 
= 0.01614. (Ans., 0.003 B.t.u.) What was the internal energy 
before expansion? (Ans., 1108.91.) After expansion? (Ans., 
800.93.) What was the external work done during expansion, 
by Equation (44)? (Ans., 307.98 B.t.u.) 

Prob. 155. What is the value of a in the exponential approx¬ 
imate formula for this process? (Ans., 1.134.) Of Ac? (Ans., 
239.) Of the external work? (Ans., 302 B.t.u.) 

Prob. 156. Consider the adiabatic expansion of Prob. 153 
as BC, Fig. 22. How much heat was absorbed, and how much 
external work was done, during the operation c6? (Ans., heat 




Steam and Other Vapors 


99 


= 305.4 B.t.u., work = 0.867 B.t.u.) Along 6J5? (Ans., heat 
= 818.04 B.t.u., work = 83.46 B.t.u.) Along Cc? (Ans., heat 
= — 777.5 B.t.u., work = — 46.37 B.t.u.) Show that for the 



closed figure cbBC, XH = 2JF. (Ans., = 305.4 + 818.04 
- 777.5 = 345.94. Also 2W = 0.867 + 83.46 + 307.98 - 46.37 
= 345.94 B.t.u.) 

Prob. 157. Ammonia is compressed adiabatically from 0° to 
90°. At 0°, n w = - 0.0709, 
n e = 1.245. At 90°, n 8 = 

1.0219. What must be the 
initial dryness if the fluid is 
to be exactly dry (x = 1.0) 
after compression? (Ans., 

0.878.) 

80. Ideal Vapor Cycle, Com¬ 
plete Expansion: “ Rankine ” 

Cycle. The cycle cbBC, Fig. 

23, represents the ideal opera¬ 
tion of a steam power plant running condensing. Water is 
heated along cb, the pressure and temperature increasing and the 
volume increasing slightly. Vaporization occurs along bB , adia- 



Fig. 23. Ideal Cycles for Vapors. 


















100 


Thermodynamics, Abridged 


batic expansion along BC and condensation along Cc. The corre¬ 
sponding diagram is similarly lettered in Fig. 22 B. We have 
seen how to compute the work and heat for each path of this 
cycle. The following is a briefer method of computing efficiency 
and mean effective pressure. Referring to Fig. 22 B, the effi¬ 
ciency is 

e = cbBC -T- gcbBD 

= (fabBD - facCD) + (fabBD - facg ) 

= (Hb — He) (Hb — h c ), 


where H indicates the total heat above 32° of steam either 
wet or dry. This is best w T ritten in the specific form, 

e = {(h+ xL )i — (A + xL) 2 } -s- {(h + xL )i — h 2 }, (45) 

in which subscripts i and 2 refer to high-pressure and low-pressure 
conditions, respectively. This efficiency is always less than that 
of the Carnot cycle between the same extreme temperatures 
The value of Xi is usually close to 1.0. That of .t 2 is computed 
from Equation (43): (n w + xn e )i = ( n w + xn e ) 2 . 

The mean effective pressure is the work in ft. lb. divided by 
{144 X ( v c — fl c )b Fig. 23. This divisor becomes x 2 (v 2 — v c ), 
or with an error practically never exceeding 1/10 of 1 per cent, 
for steam, simply x 2 v 2 , where v 2 is the tabular volume of dry steam 
at the low-pressure condition. Then 


= (*+*Di~ + x 5.403 

X 2 V 2 


(46) 


Prob. 158. Find the efficiency of a Rankine steam cycle in 
which the extreme pressures are 250 and 1 lb. and the dryness at 
the beginning of expansion is 0.99. (Ans., 0.308.) How much 
net work is done? (Ans., 345.94 B.t.u.: compare Prob. 156.) 
What is the mean effective pressure? (Ans., 7.5 lb.: mean 
effective pressures are very low in this type of cycle.) 

81. Incomplete Expansion. During the later stages of expan¬ 
sion, Fig. 23, very little work is derived from the fluid in pro¬ 
portion to the increased displacement. The cylinder of the 



Steam and Other Vapors 


101 


engine will therefore be large and costly for its power. (This 
does not apply to turbines.) In usual engine practice, expansion 
is terminated at some point E, before the pressure has been re¬ 
duced to that at which the steam is to be condensed. A constant- 
volume drop, EF, then occurs. See Fig. 22 B also. The cycle 
is cbBEF. Divide it into two areas GbBE and GEFc by a hori¬ 
zontal line. The lower area is on the pv diagram very nearly a 
rectangle, and its magnitude in B.t.u. may be taken as 


Then the efficiency is 

XW 


144 , 

yyg ^e\Pe 


Vf)' 


e = 


( h T - xL) i — h 2 


144 


(h + xL)x — (h + xL) e + ^v E (p E — p 2 ) 
(h + xL) i — h 2 


(47) 


where v E is the volume of wet vapor at the point E , to be com¬ 
puted by first finding the entropy and dryness. 

Thus under the limiting pressures and initial dryness of Prob. 
158, let p E = 10. Then at this pressure n w = 0.2832, n e = 
1 .5042. Equating the entropies at B and E, 

0.5676 + (0.99 X 0.9600) = 0.2832 + 1.5042^, 
x E = 0.819. 

Then, very nearly, v E = 0.819 X 38.38 = 31.4, the value 38.38 
being that for dry steam at 10 lb. pressure. Then the area 
GEFc = (144/778) X 31.4 X 9 = 52.1 B.t.u. The value of 
(h + xL) e is 161.1 + (0.819 X 982) = 967 B.t.u. The area 
bBEG is 375.2 + (0.99 X 826.3) - 967 = 226.24 B.t.u. Then 
2W = 52.1 + 226.24 = 278.34 B.t.u. and e = 278.34 (1193.24 

— 69.8) = 0.247. This is less than the result in Prob. 158. 
If expansion had terminated sooner, i.e., if p E had been greater 
than 10, the efficiency would have been still less. 

Prob. 159. Find the efficiency as above, when p E = 20. 

(Ans., 0.215.) 




102 


Thermodynamics, Abridged 


Prob. 160. Find the efficiency as above in the limiting case, 
when p E — 250, cycle cbBH. (Ans., 0.074.) 

The mean effective pressure with the cycle cbBEF is, very 
nearly, 7782 W -r 144^, or 


r dM (h + zL)i-(h + xL) M , 5.4032IF 

p m = 5.403- - - V Ve- V2 = —- -- • (48) 


For the foregoing conditions, this becomes 48.1. It may be as 
high as desired, up to the value pi — p 2 . Incomplete expansion 
reduces efficiency, but increases mean effective pressure. 

Prob. 161. Find p m in Probs. 159,160. (Ans., 78 and 249 lb.) 

82. Ideal Steam Consumption. The area of the cycle, or the 
numerator of the efficiency expression, represents the ideal work 
obtainable per lb. of steam, 2IF (B.t.u.). Since 1 hp. = (33 000 
X 60) -r- 778 = 2545 B.t.u. per hr., the ideal steam rate (lb. 
of steam per hr. per Ihp. under best conditions) is 2545 -f- 2IF. 
For Prob. 158, this rate is 2545 345.94 = 7.33 lb. The heat 

supplied being hi + X\L\ — h 2 , an engine having an efficiency 
of 100 per cent, would use 2545 -r- (hi + xiLi — h 2 ) lb. of steam 
per hr. For the conditions of Prob. 158, this is 2545 -f- 1123.44 
= 2.26 lb. But this ideal engine has an efficiency of only 0.308. 
Hence its ideal steam rate is 2.26 -f- 0.308 = 7.33 lb., as before. 
The actual steam consumption will be greater than 7.33, under 
the assigned conditions. 

The IDEAL HEAT RATE is the HEAT CONSUMED IN B.T.U. PER 

min. per Ihp. At 100 per cent, efficiency, it would be 33 000 
-T- 778 = 42.42. At an efficiency e it is 42.42 -r- e. In Prob. 
158, it is 42.42 -r- 0.308 = 138 B.t.u. 

Prob. 162. Find ideal steam rate and ideal heat rate in 
Art. 81. (Ans., 9.13 lb., 171 B.t.u.) 

Prob. 163. Find these ideal rates for Prob. 160. 

(Ans., 30.2 lb., 570 B.t.u.) 

Prob. 164. Find the ideal rates in Prob. 159. 

(Ans., 10.4 lb., 195 B.t.u.) 




Efficiency of Cycle 


Steam and Other Vapors 


103 


Fig. 24 shows the variation in efficiency of such cycles as 
cbBEF for steam initially dry under the initial (ps) and back 
( p F ) pressures common in condensing and non-condensing 
practice. 



Ratio of Expansion 
for non-condensing engines 


Fig. 24. Ideal Efficiencies with Incomplete Expansion: Steam Initially Dry. 
8 
















































































104 


Thermodynamics, Abridged 


Prob. 165. From Fig. 24, state the heat rate and steam rate 
for an ideal cycle using dry steam at 215 lb. pressure, condensing 
at 2 lb. absolute pressure, when the ratio of volumes during 
adiabatic expansion is 30. (Ans., e = 0.265, heat rate = 160, 
steam rate = 8.7.) 

Superheated Vapor 

83. Properties. Superheating almost always occurs at con¬ 
stant pressure. The pressure is that at which the saturated 
vapor was formed. Vapor must be dry-saturated before it can 
be superheated. The specific heat at constant pressure is 
highly variable, both with pressure and with temperature. 
Extensive tables are therefore necessary for mean values of k, 
the specific heat, or of H', the total heat. If t = saturation 
temperature, t r = actual vapor temperature, both at the pressure 
p, the “ superheat ” (amount or number of degrees of super¬ 
heat) is t' — t and the total heat is 

H' = H+ kit' - t). 

The volume is given quite closely for steam by the rather 
formidable equation 

pv’ = 0.59620' + 460) - p{ 1 + 0.0014p)(■ ( ^°^ - 0.0833 ) . 

The gain of entropy of a vapor during superheating may be 
expressed as k log e ( t ' + 460) /(t + 460), and the total entropy is 

T 

n' = n s + k log e Y * 

The external work done during superheating is 
W = ~£(v'~v) B.t.u. 

The whole amount of external work done during the formation 
of superheated steam from water at 32° and 0.0886 lb. pressure 
is (see Arts. 77, 78), 

PVio , p(v - Vv) , VW - v) vv' -n , 

5.403 U 5.403 + 5.403 5.403 t,u ‘ 







Steam and Other Vapors 


105 


The gain of internal energy during this operation is then 


w - vZ 


pv 

5^403 ’ 


Prob. 166. For steam at 250 lb. pressure, superheated 200°, 
k = 0.562. Find t', H', v', n', W' and the gain of internal energy 
from liquid at 32°. (Ans., 601.1° F.: H' = 1313.9: v' = 2.471: 
n' = 1.646, W’ = 28.7, gain of energy = 1199.7.) 

The factor of evaporation for superheated steam is 

F' = ( H ' - ho) + 970.4, 


following Equation (40). 

Prob. 167. What is the factor of evaporation for the steam 
in Prob. 166, if formed from water at 90°? (Ans., 1.29.) 

The following is a greatly abridged sample of a superheated 
steam table. 

Properties of Superheated Steam 

(Condensed from Steam Tables and Diagrams, by Marks and Davis, with 
the permission of the publishers, Messrs. Longmans, Green, & Co.) 


Superheat, °F 

40 

90 

200 

300 

Absolute Pressure Lbs. per 

r v = i4i.7 
) v' = 357.8 

191.7 

301.7 

401.7 

Square Inch 

387.9 

453.7 

513.4 

1 

1 H ' = 1122.6 

1145.3 

1195.6 

1241.5 


( n ' = 2.0069 

2.0434 

2.1145 

2.1701 


f f = 367.8 

417.8 

527.8 

627.8 


J v f = 4.72 

5.07 

5.80 

6.44 

100 

1 H' = 1208.4 

1234.6 

1289.4 

1337.8 


( n ' = 1.6294 

1.6600 

1.7188 

1.7656 


( t' = 393.1 

443.1 

553.1 

653.1 


J v' = 3.44 

3.70 

4.24 

4.71 

140 

1 H' = 1215.8 

1242.8 

1298.2 

1346.9 


( n ' = 1.6031 

1.6338 

1.6916 

1.7376 


f t' = 398.5 

448.5 

558.5 

658.5 


1 v' = 3.22 

3.46 

3.97 

4.41 

150 

1 H ' = 1217.3 

1244.4 

1300.0 

1348.8 


( n ' « 1.5978 

1.6286 

1.6862 

1.7320 


f f = 441.0 

491.0 

601.0 

701.0 


v' = 1.98 

2.14 

2.47 

2.75 

250 

H' = 1229.3 

1257.7 

1313.9 

1363.5 


l n' = 1.5593 

1.5900 

1.6460 

1.6905 











106 


Thermodynamics, Abridged 


84. Rankine Cycle with Superheat. Such a cycle is repre¬ 
sented as cbdxJ, Fig. 22 B, or as cbBC, Fig. 23. There is no 
change in direction of the constant pressure line on the pv 
diagram. The efficiency is 


Thus, take the conditions as p x = 250, t x = 601°, p c = 1. 
First find the dryness at J : 

( n w + xn e )j = n' = 1.646 (Prob. 166) 

0.1327 + 1.8427*, = 1.646, z, = 0.82. 

Taking from Prob. 166 H x = 1313.9, 


1313.9 - {69.8 + (0.82 X 1034.6)} 
1313.9 - 69.8 


0.319. 


In Prob. 158, the efficiency for the same pressures but with 
slightly wet steam was 0.308. Superheating improves 

EFFICIENCY. 

The mean effective pressure, following Equation (46), is 
5.4032 IF 

Vm ~ (xv)j 

Noting that 2 W is the numerator of the expression for efficiency, 
this becomes (5.403 X 397) -f- (0.82 X 333) = 7.84 for our 
conditions. 

Prob. 168. Find e and p m in Art. 84 if the superheat is 100° 
instead of 200 °. (Ans., e = 0.313, p m = 7.64.) 

85. Incomplete Expansion. Following Art. 81, if expansion 
terminates before the line of back pressure is reached, the work 
of the superheated cycle is 

XW = J/; - (h + xL ) 3 + , 

in which subscripts 3 refer to the “ terminal ” condition, E, 
Fig. 23. The cycle is cbBEF ( cbdxKF in Fig. 225). The 
efficiency is 

2 IF 

H' x -h c 


e = 


(50) 







Steam and Other Vapors 


107 


and the mean effective pressure is 


Pm — 


5.40321V 

% 


(51) 


Suppose that under the conditions of Art. 84 the terminal 
pressure is p K = p 3 = 10. Then 


nj = 1.646 = (n w + xn e ) 3 = 0.2832 + 1.5042ar 3 , 
# 3 = 0.906, ( xL) z = 0.906 X 982 = 891, 

v K = 0.906 X 38.38 = 34.9 = i> 3 , 


34 Q V Q 

21V = 1313.9 - 161.1 - 891 + - J * = 320 B.t.u., 

5.403 

n 5.403 X 320 „ 

— 0.257, p m — 34 9 — 49.6. 


e = 


320 


1313.9 - 69.8 


The efficiency exceeds that computed for saturated steam in 
Art. 81. 


Prob. 169. Find e and p m when the pressure limits are 250 
and 1 lb., the initial superheat 100° and the terminal pressure 
20 lb. (Ans., 0.223 and 79 lb.) 

86 . Cases Not Hitherto Considered. In Arts. 84 and 85, the 
steam at the end of its adiabatic expansion was wet. In cases 
where at this point it is still superheated, an exact solution is 
not possible by any method thus far shown. We know in general 
the terminal pressure and entropy. We have no way of finding 
terminal volume, temperature, superheat, or heat content. 
A graphical method for finding these properties will now be 
developed. 

In Fig. 25 the coordinates are total heat above 32° and total 
entropy. Plot the “ saturation curve ” ab from values in the 
steam table. Thus at 100 lb. pressure the coordinates are 
1186.3 and 1.602: at 1 lb. they are 1104.4 and 1.9754. These 
values establish the points b and n. 

The points e and c are those at which the pressures are 150 
and 50 lb. Plot the constant pressure curves ef and cd. 
This is done by finding the coordinates of points like /, for some 






10S 


Thermodynamics, Abridged. 


assumed dryness. Thus, assume a dryness of 0.80. Then at / 
the coordinates are, 330.2 + (0.8 X 863.2) = 1020.8, and 0.5142 
+ (0.8 X 1.055) = 1.3582. Taking other drynesses, additional 
points may be located to establish the whole line ef. Plot cd in 
the same way. Joining / and d gives a line of constant dryness. 



Prob. 170. Locate the point o on the line of 1 lb. pressure, 
where the dryness is 0.80. (Ans., coordinates are 897.5 and 
1.6069.) 

The constant-pressure curves are now carried upward into 
the superheated region, using the table in Art. 83, and lines of 
CONSTANT SUPERHEAT, gj, hk, are drawn. 

Prob. 171. Find coordinates of points g and j. (Ans., 1244.4 
and 1.6286: 1145.3 and 2.0434.) 

87. Application of Total Heat Entropy Diagram. For com¬ 
plete expansion (Rankine) cycles, Equations (45) and (49) 
may both be written 

















































































Steam and Other Vapors 


109 



(52) 


where subscripts i and 2 refer to conditions at the beginning 
and end of adiabatic expansion, respectively, h stands for heat 
of liquid and Q for total heat of vapor either wet, dry or super¬ 
heated. Values of Qi and Q 2 , and also of the final dryness, x 2 , 
may be read directly from a chart like Fig. 25. A more complete 
chart appears in Fig. 26. Still larger scale charts, which may 
be used for accurate work, will be found in recent steam tables 
like those of Goodenough or Marks and Davis. The saving in 
labor due to their use is very great. Thus suppose a cycle with 
complete expansion to work between pressure limits of 150 lb. 
and 1 lb., with an initial superheat of 200°. Locate h, Fig. 25 
and read off Qi = 1300 at the left. Draw hi vertically. Read 
off, at l, interpolating between the 0.84 and 0.85 dryness curves, 
x 2 = 0.842, Q 2 = 944. Thene = (1300 - 944) 4- (1300 - 69.8) 
= 0.256. The value h 2 = 69.8 is from the steam table. Taking 
the specific volume at 1 lb. pressure at 333, also from the steam 
table, 

5 403 X 356 

= 0.842 X 333 = 280.4, y m = ' -= 6.86. 


This method is applicable whether the steam is initially wet, dry 
or superheated. 

Incomplete Expansion. The most difficult case is the one thus 
far omitted, that in which the steam is superheated at the end 
of expansion. Equations (47) and (50) may be written 


Qi — Qs + 


VsiVs ~ P2) 

5.403 


Qi h 2 


(53) 


where 3 is the terminal condition. If this condition is one of 
wet steam, the volume v s is found from the dryness, as above. 
But suppose steam at 150 lb. pressure, 200° superheated, to 
expand to yz = 50 lb. pressure and then to be condensed at 
1 lb. pressure. In Fig. 25, Qi = 1300, as before: draw hm 






110 


Thermodynamics, Abridged 


o lo o lo o lo o lo o lo-olooloolooloOloolooloOlooloo 
lo n o n to n o n w oJOf'^ir>cJor'-Ln£f2r—ir>cvjor^.LOc'JOi v -Ln 

•'T Tj-r^rocororoCMCJ OJ OJ — — — — OOOO^^ocDOOOOooOOr^r^ 



LO O LO O LO O LOOLOOLpOLOOmOlD OUOOLOO LO O LO O LO O LO O 

r^Locvic>r~'io ojor-'Loc>Jor~'LOfvior-~Loojc>r— ioruor~~LO(vioo~Ln 
^2:2t5l£2£2£12£2£i!^i£^ — z z z z 2 2 2 2 <J) o> a) a) co co oo oo i^ 
'D-V9‘JoZe ©Aoqe ^bsh leq-oi 


Entropy above 32°F. 

Fig. 26. Total Heat-Entropy Digaram. 












































































































































































































































































































Steam and Other Vapors 


111 


vertically. At m, Qz = 1203, superheat = 54°. Then U = 281 
+ 54 = 335, and the equation in Art. 83 for the volume of super¬ 
heated steam gives vz = 9.26. Then 


e = 


1300 - 1203 + (50 - 1) 


and 


1300 - 69.8 

181 X 5.403 


181 

1230 


= 0.147 


Pm — 


9.26 


= 106. 


Note the marked difference between these results and those of 
the last paragraph. Incomplete expansion has reduced the 
efficiency from 0.256 to 0.147 but has increased the mean 
effective pressure from 6.86 to 106. A corresponding increase 
would occur in the power of an engine of given size and speed. 


Prob. 172. Check from Fig. 26 all drynesses and efficiencies 
computed in Probs. 153, 158-160, 168, 169. 

To avoid the labor of computing u 3 from the long formula of 
Art. 83, an auxiliary chart called the total heat-pressure 
diagram is often used. This uses the quantities mentioned as 
coordinates, and plots curves of constant dryness, superheat and 
volume. Pressure and heat content being taken from Fig. 26, 
volume may be read directly from the total heat-pressure 
diagram. 

Factors Affecting Ideal Engine Efficiency 
88. Incomplete Expansion, Dry or With Superheat. Effi¬ 
ciencies of various cycles with superheat are plotted in Fig. 27. 
These may be compared with values in Fig. 24. 

Back Pressure has of course a pronounced influence. This 
explains the gain in economy by running condensing. A very 
low ratio of expansion, running condensing, gives better efficiency 
than a high ratio running non-condensing. A back pressure 
around 25 lb. absolute is often used for small auxiliary engines 
on ships. This is for the purpose of using the exhaust steam 
from these engines advantageously in heating the boiler feed 
water. Fig. 24 shows that it leads to low engine efficiency. It 




Ideal Thermodynamic Efficiency 


112 


Thermodynamics, Abridged 


pays, because practically all of the heat wasted by the engines 
goes back to the boilers. 


Ratio of Expansion-Condensing Engines 
5 10 15 20 25 30 35 



Fig. 27. Ideal Efficiencies with Superheated Steam. 


















































Steam and Other Vapors 


113 


Ratio of Expansion has an especially important bearing on 
efficiency. Many land engines running at constant speed are 
governed by varying this ratio. Hence the curves show how 
efficiency varies with output. The best ratio is not to be deter¬ 
mined by thermodynamic considerations alone, but is established 
from various practical and commercial factors (Arts. 95, 96). 
At a given pressure, efficiency increases with the ratio of expan¬ 
sion, but mean effective pressure decreases. A ratio is finally 
reached at which there is little or no further gain in efficiency. 
This limit is very soon attained if the initial pressure is low, and 
is postponed if high superheat is used. 

High Initial Pressure is important. At a given ratio of ex¬ 
pansion, the efficiency increases with the pressure: but for 
very low ratios, high pressures are of little advantage. In con¬ 
densing engines with superheat, high pressure is less important 
than in non-condensing engines using superheat. 

Superheat increases efficiency at a given pressure or ratio of 
expansion. In general, the gain of efficiency is roughly pro¬ 
portional to the amount of superheat. Superheat should not be 
continued beyond the point where the terminal condition is 
dry. Superheat gives high efficiency at moderate ratios of 
expansion and therefore permits of reasonably high mean 
effective pressures along with good efficiency. 

Initial Dryness is not considered in these diagrams. Within 
reasonable limits (0.95 to 1.0) it has practically no influence on 
efficiency. 

All efficiencies are below those of the Carnot cycle between the 
same extreme temperature limits. 

89. Vapors Other Than Steam. In the Carnot cycle, the 
efficiency is independent of the fluid used. In the actual steam 
cycles here described, it depends on the properties of the fluid 
(latent heat, specific heats). Various fluids have been used in 
vapor engines. In simple condensing engines with low ratios 
of expansion, a considerable gain is possible over the best results 
with steam. A desirable vapor would be one whose pressure 


114 


Thermodynamics, Abridged 


was less than that of steam at high temperatures and more at 
low temperatures. The latter characteristic w T ould make high 
vacuums unnecessary. It should have a high latent heat of 
vaporization and a low specific heat of liquid, since this, in 
Fig. 22 B, would make the Rankine cycle approximate more 
closely the Carnot. In the cycle cbBC of this diagram, the heat 
supplied by the boiler is the area under cbB. That carried 
away at the condenser is the area under Cc. The ratios of these 
areas to the power developed depend on the properties of the 
fluid used. Hence a suitable substitute fluid might decrease the 
relative size of boiler or condenser necessary. The mean effective 
pressure, and hence the size of cylinder, also depends on the 
properties of the vapor. 

Prob. 173. In Fig. 20, what vapor apparently has a higher 
pressure than that of water at low temperatures, and a lower 
pressure at very high temperatures? 

Mixtures of Air and Steam 

90. Saturated Air. The word “saturated,” as applied to air, 
has a different meaning from that of the same word applied to 
steam. Saturated steam is dry steam. Saturated air is the 
wettest kind of air. Supersaturated air is air containing all 
the moisture it can hold, in the presence of an additional body 
of moisture. 

When air and a sufficiency of moisture are mixed at the 
temperature t, the mixture contains water vapor, i.e., steam, 
at the corresponding pressure p s , which may be taken from the 
steam table. The total pressure of the moist air is the sum of 
the partial pressures of dry air and steam. If we are dealing 
with normal barometer, the partial pressure of dry air is then 
p a = 14.696 — p s . The steam is saturated steam and the air. 
is saturated air. One cubic foot of pure dry air unmixed with 
steam at the temperature t and normal barometer would weigh 
(144 X 14.696) ~r~ (53.36 T) lb. One cubic foot of air in the 
present saturated mixture weighs w a = 144(14.696 — p s ) -f- 
(53.36 T) lb., which is less than the weight of dry air unmixed 


Steam and Other Vapors 


115 


with steam. The weight of steam intermixed with the 1 cu. ft. 
of dry air in the mixture is w s = 1/v, where v is the specific 
volume from the steam table at t° F. By Dalton’s law, if we 
employ partial pressures p s and p a as above, 1 cu. ft. of mixture 
contains both 1 cu. ft. of dry air and 1 cu. ft. of steam. 

Prob. 174. What is the weight of a cubic foot of pure dry air 
at normal atmospheric pressure and 40° F.? (Ans., 0.0794 lb.) 

Prob. 175. The pressure of saturated steam at 40° F. is 
0.1217 and its specific volume is then 2438. If air at 40° F. and 
normal barometer is saturated with moisture, state the partial 
pressures of steam and air. (Ans., 0.1217 and 14.5743.) 

Prob. 176. In Prob. 175, state the weights per cu. ft. of air 
and steam. (Ans., 0.0787 and 0.00041 lb.) What is the density 
(weight per cu. ft.) of the mixture? (Ans., 0.07911 lb.) 

91. Unsaturated Air. The “absolute humidity” is the ratio 
of weight of steam to weight of air. In Prob. 176, it is 0.0052. 
If there is insufficient moisture present to saturate the air, its 
temperature will be t, but its pressure will be p s r , less than p s . 
Hence the moisture in such a mixture will be superheated 
steam. Call its weight per cu. ft. w s '. Then the relative 
humidity is defined as r h = p s '/Ps = w s '/w s . The partial pres¬ 
sure of air in the mixture, at normal barometer, is 14.696 — p 8 ', 
from which the weight of dry air in a cubic ft. of mixture may be 
computed as before. The weight of steam will be ThW s . 

Prob. 177. What is the absolute humidity of saturated air 
at 212° F.? (Ans., ».) 

Prob. 178. At t = 200°, p s = 11.52, w s = 0.02976. For a 
relative humidity of 0.9, find for moist air the values of p 8 r 
(Ans., 10.368), pd (Ans., 4.328), w 8 r (Ans., 0.026784), w a ' (Ans., 
0.0177). 

Prob. 179. As t increases, how will p s , p a , w 8 and w a vary? 
(Ans., p 8 and w s will increase, p a and w a will decrease.) 

92. Moist Air Processes. Following Equation (13), the value 
of R for a mixture of air and steam is 


116 


Thermodynamics, Abridged 


Rm = 


R s w s ' + 53.36 w a ' 85.8 w* + 53.3 Qw a ' 


W s ' + Wa 


Ws' + Wa 


the value of R for steam being taken at 1544/18.016 = 85.8. 
R m will often differ notably from 53.36. 

When a fixed body of unsaturated moist air is cooled, the value 
of w s ' does not change, but that of w s decreases. Hence the 
relative humidity increases. If the process is carried sufficiently 
far, the mixture will become saturated air. Beyond this point, 
the air will be supersaturated and dew will be deposited. Sup¬ 
pose Th = 0.60 at 100° and the mixture is cooled to 90°. The 
original 1 cu. ft. becomes (460 + 90)/(460 + 100) = 0.982 
cu. ft. At 90°, the density of saturated steam is 0.002131. The 
weight of steam in 0.982 cu. ft. of mixture, if saturated, would 
be (w s ) 9 o = 0.982 X 0.002131 = 0.00209. At 100°, the weight 
present if saturated is 0.002851. This weight, multiplied by 
the original relative humidity of 0.60, is still present after cooling 
to 90°. Hence at 90° the relative humidity will have become 


r h 


fw/\ 0.002851 X 0.60 
\ w s / 90 0.00209 


0.82. 


Prob. 180. Find the value of R m in Prob. 176. (Ans., 53.5.) 

Prob. 181. In Art. 92, what is the relative humidity if the 
mixture is further cooled to 85°, where iv s = 0.001832? (Ans., 
0.96.) 

Pure steam at 102° F. has a pressure of 1 lb. absolute. Hence, 
however low the temperature of condenser circulating water, 
a low temperature of heat rejection in the steam engine (shown 
in Art. 36 to be essential for highest potential efficiency) is 
dependent on the maintenance of a good vacuum by the air 
pump. By using a mixture of air with steam in an engine, a 
rejection temperature as low as the circulating water will permit 
might be attained along with any pressure or vacuum (above the 
pressure given in the steam table for cooling water temperature) 
which might be desired. We might have the steam and air 
at 70° F. The partial pressure of the steam would then be 





Steam and Other Vapors 


117 


only 0.3626 lb. per sq. in., with saturated air. The total pressure 
might however be 1 or 2 lb., a range of values easily maintained 
by modern vacuum apparatus. 

Prob. 182. A mixture of air and steam at 2 lb. pressure has a 
temperature of 126.15° F. What would be the air pressure if 
the air were saturated? (Ans., 0.) What are the steam and 
air pressures if th — 0.5? (Ans., each 1 lb.) In the latter case, 
how much is the steam superheated? (Ans., 24.32°.) In a cubic 
foot of such mixture, state the weights of air and steam. (Ans., 
0.00461 and 0.00289 lb.) The mixture is cooled at constant 
pressure to 70° F. What will then be its volume? (Ans., 0.903 
cu. ft.) What weight of steam could be held by this volume of 
mixture, if saturated? (Ans., 0.001037 lb.) Is it saturated? 
(Ans., super-saturated.) How much steam must have been con¬ 
densed as dew? (Ans., 0.001853 lb.) What weights of water, 
saturated steam and dry air are now present? (Ans., 0.001853 
lb. water, 0.001037 lb. dry steam, 0.00461 lb. air.) How much 
heat above 32° is contained in the water and steam? (Ans., 
1.2005 B.t.u.) 


CHAPTER V 


EFFICIENCY AND POWER OF STEAM ENGINES 


Efficiency 

93. Prediction of Efficiency. Following Art. 82, if an engine 
uses W s lb. of steam per hr. while developing M Ihp., its actual 

INDICATED THERMAL EFFICIENCY is 


2545 


2545M 




w 


W a (Q i - A,) 


(54) 


where Qi = total heat above 32° of steam at the throttle condi¬ 
tion (wet, dry or superheated) and h 2 = heat of liquid at the 
pressure existing in the exhaust pipe. Let e denote the efficiency 
of the Rankine (complete expansion) cycle under the conditions 
Qi, h 2 . Then the relative efficiency, or ratio of actual and 
ideal efficiencies, is 

e R — e a “ ( 55 ) 

The heat rate of the actual engine is 
W s 

wii ^ ~ ^ per m ^ n * 

The work done by it per lb. of steam is 

/ r\ 7 \ 2545 

e a {Q i - h 2 ) = w * ^ ^ B.t.u. 

Prob. 183. Under the conditions of Prob. 158, a certain 
compound engine uses 13.3 lb. of steam per Ihp.-hr. Find the 
actual efficiency, relative efficiency, heat rate and work per lb, 
of steam. (Ans., 0.17: 0.55: 250: 190.) 

94. Probable Values. A rough estimate of the probable 
economy of a new engine for definite steam conditions may be 
formed by using the following values of e R \ 

118 





Efficiency and Power of Steam Engines 


119 



Simple 

Compound 

Triple 

Expansion 

Saturated steam, non-condensing . . 

0.6 

0.65 


Saturated steam, condensing. 

0.4 

0.5 

0.6 

Highly superheated steam. 

0.65 

0.65 



Best Recorded Performances, Standard Engines 


Saturated Steam, 
Lb. Per Ihp.-hr. 


Simple non-condensing.21 

Simple condensing.\..16.3 

Compound non-condensing.16 

Compound condensing.12 

Triple-expansion condensing.10| 


Steam Superheated 
Above 150° F., 
B.t.u. per Ihp.-min. 

300 

226 

245 

200 

190 


Records as good as these are rarely attained. The following 
steam rates may be realized with saturated steam by engines in 
good condition and operating at their most favorable loads: 



Single 

Valve 

Double 

Valve* * 

Four 

Valvet 

Four Valve, 
Releasing! 

Simple non-condensing. 

33 

30 

29 

26 

Simple condensing . 

27 

23 

2H 

21| 

Compound non-condensing. 

25£ 

23 

22 

22 

Compound condensing. 

20 

17 

15 

15 


These rates will be surpassed with very high steam pressures 
or exceptionally good vacuum. 

95. Indicator Diagram vs. Rankine Cycle. So far as efficiency 
is concerned, the Rankine cycle, which represents the operation 
of the whole power plant, establishes a limit for the engine 
specifically. A still closer limit is found in the “ incomplete 
expansion ” cycle, ABCDF, Fig. 28. The actual diagram from 
the engine appears relatively as abceg. The rounding of corners 
in the actual diagram is due to slow movements of valves. The 
reduced admission pressure and augmented back pressure are 
caused by friction of steam flowing through valve ports and 

* Separate cut-off control. 

f Admission, cut-off, release and compression independently controlled. 

+ Automatic disengagement causing very sharp cut-off. 

9 






























120 


Thermodynamics, Abridged 


passages. Clearance exists as a necessary evil in the actual 
engine. It is relatively large in engines of small size. 

The operation of compression, eg, is chiefly introduced for 
its cushioning effect, to promote quiet and smooth running. 
Compression is greatest in high-speed engines. From an eco¬ 
nomical standpoint its duration should probably never exceed 
10 per cent, of the stroke. 


^Pressure at Throttle g ^ 


Ratio of Expansion=^Lor^\ 



'Pressure of Atmosphere or Condenser 


Fig. 28. Action of Steam Engine. 


The lines be and BC are unlike in slope and position. In 
particular, the volume at b is much less than that at B. This is 
due to cylinder condensation, an important factor in engine 
economy. If the cylinder were covered with a perfect heat- 
insulating material, there would nevertheless be a rapid alterna¬ 
tion of temperature of its metal walls, according to the tempera¬ 
ture of the steam in contact with them. The walls are heated 
during admission and the early part of expansion and cooled 
during the late expansion and exhaust. They abstract heat from 
the incoming steam very rapidly, so that at the point of cut-off 
the steam is often less than 0.60 dry. This accounts for the 













Efficiency and Power of Steam Engines 


121 


small volume at b. Late in the expansion stroke, the steam has 
cooled off and the walls of the cylinder now supply it with heat. 
This raises the right-hand end of the expansion curve so that 
be is less “ steep ” than BC. The restoration never compen¬ 
sates for the initial loss. The availability of heat for doing 
work depends on its temperature (Art. 36). Low-temperature 
heat is worth less than the same quantity of high-temperature 
heat. 

The curve be is therefore not adiabatic. It may be represented 
by an equation pV a = const., a rough average value of a being 
1 .0. This does not make it isothermal, the saturated steam 
isothermal being a line of constant pressure. For non¬ 
condensing unjacketed engines, a = (x b + 0.614) -r- 1.258, nearly. 

Condensation is accentuated by high ratios of expansion, 
which increase the magnitude and duration of the temperature 
alternations. It is apt to be excessive at slow speeds. It is 
large in small engines, because their ratio of surface to volume 
is large. It is reduced in multiple-expansion engines (Art. 
100 ). 

Prob. 184. What steam consumption might be expected from 
the tabular value of e R in a simple condensing engine under the. 
conditions of Prob. 183? (Ans., 18.1.) How much better than 
this was the performance of the best standard simple condensing 
engine? (Ans., 10 per cent, better.) What is the chief reason 
why an average good engine uses 21^ lb.? (Ans., it uses a lower 
steam pressure.) 

Prob. 185. In Fig. 28, will the rounding of corners be most 
pronounced at light loads or heavy loads? Will the differences 
of pressure between ab and AB, ce and DF, be greater at slow 
speeds or at high speeds? Why should compression be greatest 
in high-speed engines? 

Prob. 186. What is the value of x b when that of a for the curve 
be, Fig. 28, is 1.0? (Ans., 0.644.) What then is the approximate 
value of V b -s- V B ? (Ans., 0.644.) What is a when x b = 1.0? 
(Ans., 1.28.) 


122 


Thermodynamics, Abridged 


Prob. 187. Assume that condensation varies inversely with 
speed. The condensation loss is 10 per cent, at 300 r.p.m. 
What will it then be at 100 r.p.m? (Ans., 30 per cent.) 

Prob. 188. The surface of a cylinder is tv dL + (x/2)d 2 
= Tvd(L + (d/2)) and its volume is (7r/4)d 2 Z. The ratio of 
surface to volume is therefore 4/d + 2/L. If the actual con¬ 
densation loss is proportional to surface and the weight of steam 
used is proportional to volume, state a ratio expressing relative 
condensation loss per lb. of steam used. (Ans., 4/d + 2/L.) 
Compare condensation losses in 10 by 20 and 40 by 60 in. cyl¬ 
inders, all other factors being disregarded. (Ans., the condensa¬ 
tion loss per lb. of steam in the large cylinder is only 26.7 per 
cent, of that in the small.) 

96. Ratio of Expansion. Values used for ratio of expansion 
in practice are less than those of highest ideal efficiency, because 
of condensation. We are limited by condensation to com¬ 
paratively low expansion ratios. Values from 3 to 5 are common 
in simple engines, and up to 36 in compounds. Too high a ratio 
leads to large size of cylinders and excessive cost (p m being low) 
as well as to excessive condensation. In constant speed engines 
the ratio of expansion decreases with increased power. The 
efficiency increases with increase of power up to about the 
designed load (Fig. 29). The engine should give its best effi¬ 
ciency at the load usually carried. Fig. 29 gives several curves 
of steam rate against output or ratio of expansion. All show 
poor economy at high ratios (light loads). Practically all show 
poor economy at very low ratios (heavy loads). The low points 
of the curves are the best operating points. A flat curve is 
particularly desirable where the load is variable. 

97. Jackets. The steam jacket consists of a hollow space 
surrounding the cylinder, filled with high-pressure steam. By 
keeping the cylinder walls hot at all times, it prevents or dimin¬ 
ishes condensation. The steam condensed in the jacket is of 
course chargeable to the engine (Qj — /i 4 B.t.u. per lb., where 
Qj = total heat above 32° of steam supplied, h± = heat of liquid 


Efficiency and Power of Steam Engines 


123 


at the temperature of the jacket drip), but there is usually a 
net gain. Values of ■ tabulated in Art. 94 may be increased 
0.03 to 0.05 if jackets are employed. The jacket is typical of a 
large class of devices which permit of the saving of fuel by the 
expenditure of money. It is therefore to be used where fuel 

34 


Q- 

fee 

CD 

CL 

=S22 

a; 

3 18 

£ 

<0 I. 

14 

LO 

10 


Fig. 29. Variation of Economy with Ratio of Expansion: Constant Speed 
Engine. 

saving is most important: that is, where fuel is costly, or the 
space which it occupies is valuable, or the load steady. The 
maximum percentage gain due to jacketing will be realized if 
the ratio of expansion is high and the speed low. The reason is 
that these conditions would lead to excessive cylinder condensa¬ 
tion if jackets were not used. Compression may be especially 
small in jacketed engines. 

Prob. 189. In Fig. 29, state the per cent, of rated hp. at 
which best economy is attained, in engines A and B. (Ans., 
90 and 103.) 

Prob. 190. The Rice and Sargent engine in Fig. 29 was rated 
at 700 hp. What weight of steam would it use per hr. when 
developing 490 hp.? (Ans., 6700 lb.) 
















































sSsc 




















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A 






X 




l 

b'nc 

line 




A 




\T 















































R- 






°nsmg 













D 























\ 



1 


















Fro 

>m , 

sar 

rip enr. 

r/nt 

3 












S i. 





Co/ 

nripn c/n/7 

























—J 

K ^ 



















\ 




















N 


7/?Oi 

7SP- 



















it 






















—1^1 

A/c 




















cr dr yd/ c/cr/// lu/ uyuui /v 



40 50 60 70 80 90 100 110 120 


Percent of Rated hp. 

High^—Ratio of Expansion ^-Low 





































































































124 


Thermodynamics, Abridged 


Prob. 191. Which engine in Fig. 29 has the lowest steam rate 
at normal load? Which at 70 per cent, of normal load? At 125 
per cent.? 

Prob. 192. Under the conditions of Prob. 158, an engine 
uses 12 lb. of cylinder steam and 1 lb. of jacket steam, both per 
Ihp.-hr. The jacket supply is dry steam at 240 lb. pressure and 
its drain temperature is 386°. How many B.t.u. per hr. are 
chargeable as cylinder steam? (Ans., 13481.2.) As jacket 
steam? (Ans., 841.7.) Total? (Ans., 14322.9.) What is the 
actual efficiency? (Ans., 0.178.) The steam rate? (Ans., 13 
lb.) The relative efficiency? (Ans., 0.578.) 


98. Superheat. The thermodynamic advantage of superheat 
is suggested by Fig. 27. In practice, the steam which expands 
in the cylinder is usually not superheated, unless superheat 

exceeding 100° is provided at 
the boiler. The reason is 
that the cold cylinder walls 
cool the steam during admis¬ 
sion. Superheat below 100° 
or 150° simply mitigates con¬ 
densation. Higher superheat 
gives the additional advantage 
of a more efficient cycle. In 
Fig. 28, if ABCDF is the ideal 
cycle and abceg the actual 
cycle realized with saturated 
steam, superheat (even slight 
superheat) moves the point b 
toward L and gives an ex¬ 
pansion curve like LM, “ flat¬ 
ter ” than the curve be. Since 
with superheat we should use a larger ratio of expansion, the 
actual curve, representing the behavior of reduced weight of 
steam, lies like b'c'. The mean effective pressure is then less 
when superheat is used. Fig. 30 shows the effect of low super- 



Fig. 30. Effect of Low Superheat 
(72° to 92°) on Overall Economy. H. 
M. S. Brittania. 





































Efficiency and Power of Steam Engines 125 

heat. Fig. 31 gives average results at low and high superheat 
from tests of 40 famous engines. Superheat exceeding 200° 
may be expected to produce 
a saving of 20 per cent, in 
simple engines and 16 per 
cent, in compounds. The 
simple engine with super¬ 
heated steam is often as 
economical as the compound 
with saturated steam. The 
conditions which make su¬ 
perheat desirable are the 
same as those which favor 
the use of jackets (Art. 97). 

The two should not be used 
together. Low compressions 
should be used when superheat is employed. 

Prob. 193. From Fig. 31, what is the gain by using 400° 
superheat as compared with 100°? (Ans., 15 per cent.) 

Prob. 194. In Fig. 30, what was the load of best economy? 
(Ans., 70 per cent, of full load.) With saturated steam, how 
much more total coal is used per hr. at full power than at best 
economy? (Ans., 54 per cent.) Note that in warships engines 
are designed for best economy at reduced power and speed. 

99. Speed. If an engine duplicated its own indicator diagram 
continuously, the mean effective pressure would be constant, 
and the ihp. would be directly proportional to the r.p.m. In 
single-valve engines with fixed cut-off, the valve action at high 
speeds is inferior. Hence the amount of steam used tends to 
increase somewhat more rapidly than the ihp. This is just about 
offset by the favorable influence of high speed on cylinder con¬ 
densation. The higher the speed, the less time is available 
for transfer of heat. From this standpoint, the steam consump¬ 
tion should increase a little less rapidly than the Ihp., as the 
speed increases. As a final result, when the total hourly steam 



Fig. 31. Relation Between Efficiency 
and Superheat. Compound Condens¬ 
ing Engine. 























126 


Thermodynamics, Abridged 


consumption is plotted against Ihp., the resulting curve is (for 
engines of this type) a straight line. The steam rate (lb. per 

Ihp.-hr.) is therefore best 
(lowest) at high speeds. 

In marine engines, with re¬ 
versing valve gears, cut-off is 
delayed* at high speed, which 
is the full-power condition. 
The power therefore increases 
faster than the speed (Fig. 
32). The most economical 
point of cut-off, and of the 
ratio of expansion, are fixed 
at maximum speed for mer¬ 
chant vessels and at reduced 
cruising speed for warships. 
For the former, the steam 
rate per Ihp. hr. should de¬ 
crease as the speed increases. 
For the latter, it should vary somewhat like the coal rate of 
Fig. 30. 



Fig. 32. Indicated Power of an Ar¬ 
mored Cruiser at Various Speeds. 


Prob. 195. An auxiliary engine uses 1730 lb. of steam per hr. 
when developing 100 hp. and 11 730 lb. when developing 500 hp. 
The curve of hourly steam against hp. is a straight line. What 
is the steam rate when developing 300 hp.? (Ans., 22.43 lb. 
per hp.-hr.) 


Prob. 196. In Fig. 32, compare mean effective pressures at 
90 and 120 r.p.m. (Ans., p m at the low speed is 51 per cent, of 
that at the high speed.) 


100. Multiple Expansion. By partially expanding the steam 
in one cylinder and then conducting it to another, the alternations 
of temperature in any one cylinder are reduced. Hence con¬ 
densation is diminished,* and a much larger total ratio of ex- 


* The temperature range in any cylinder is halved, but the total size 
(surface) of cylinders is not doubled. 







































Efficiency and Power of Steam Engines 


127 


pansion may be utilized with advantage. . These ratios range up 
to 30 or more in compound (two-stage) and triple expansion 
engines. They are rarely over 5 in simple engines. There is a 
corresponding gain in economy. The compound will have a 
steam rate 20 to 30 per cent, lower than the simple engine. 
The triple may be slightly better still. Four stages (quadruple 
expansion) are scarcely to be considered for naval service. The 
number of cylinders is often greater than the number of expansive 
stages, in order to avoid the use of very large low pressure 
cylinders. Compound engines usually have small compression. 
They should always have high initial pressure and should run 
condensing. The reheater is a vessel placed between the 
cylinders. Steam passes through it on the way from one cylinder 
to the next. It is equipped with coils containing high-pressure 
steam. This dries and often superheats the steam which is on 
its way to the second stage. The heat supplied is computed in 
the same way as the heat supplied at a jacket. The argument 
in favor of the use of a reheater is the same as that for the use 
of superheat in any cylinder. 

Prob. 197. From the last table in Art. 94, state the per¬ 
centage gain by compounding in four valve engines, ( a ) con¬ 
densing, (6) non-condensing. (Ans., ( a ), 30: (6) 24.) 

101. Quality of Exhaust Steam. The heat supplied to an 
engine, per lb. of steam, is Qi — h 2 : Art. 93. The work done is 


2545 
W 8 + M 


2545 X Ihp. 

W s 


B.t.u. per lb. of steam. 


The heat rejected is the difference of these two quantities. 
It is equal (ignoring radiation) to ( xL ) E , where the subscript E 
refers to the condition of steam in the exhaust pipe. Then 


{xL) e — Qi h 2 jl. m' 

These relations are important in condenser design. The con¬ 
denser may however absorb more heat than that indicated by 
Equation (56), because it may cool the condensation. 





128 


Thermodynamics, Abridged 


Prob. 198. In Prob. 183, how much of the heat in each lb. 
of steam is rejected to the condenser? What is the dryness of 
the steam in the exhaust pipe? (Ans., 933.44 B.t.u.: 0.90.) 
If the condenser cools the condensed steam to 90°, how much 
additional heat does it remove? (Ans., 11.8 B.t.u. per lb.) 

102. Desirable Back Pressure. Even from a purely thermo¬ 
dynamic standpoint there is a limit to the condenser vacuum 

desirable in simple engines. 
In Fig. 33, fbcde is an in¬ 
complete expansion cycle. 
Consider the effect of lower¬ 
ing the back pressure, so 
that the cycle becomes 
abcdg. The gain of work is 
afeg. The increased cost of 
heat is hf — h a . The value 
of afeg is from the pV dia¬ 
gram, practically, (® e /5.403) 
ip/ — p a ) B.t.u. Now for 
engines of any importance, v c 
is never greater than 4. This volume is the volume of dry (or 
nearly dry) steam at the throttle pressure. In simple engines, the 
ratio of expansion is practically never over 5. Now v e is equal 
to v c multiplied by the ratio of expansion. Hence a maximum 
value of v e is 20, and a maximum value of afeg is 3.7 (pf — p a ), 
in simple engines. If p/ = 2, p a = 1, afeg = 3.7 B.t.u. But, 
then, hf = 94, h a = 69.8, heat cost = 94 — 69.8 = 24.2 B.t.u. 
The efficiency at which we have gained the additional area afeg is 
then 3.7 -f- 24.2 = 0.152. This may easily be less than the 
efficiency of the original cycle fbcde : hence the lower back 
pressure (improved vacuum) will lower the efficiency of the 
whole cycle. Simple engines should operate at 2 lb. back 
pressure (26 in. vacuum). With compound engines, the ratio 
of expansion is much greater and a better vacuum is desirable. 

Prob. 199. A compound engine uses dry steam at 250 lb. 
pressure and a ratio of expansion of 30. Find v e . (Ans., 55.5.) 







Efficiency and Power of Steam Engines 


129 


How much work is gained per lb. of steam by lowering the back 
pressure from 2 lb. to 1 lb.? (Ans., 10.3 B.t.u.) At what 
efficiency is this quantity of work derived? (Ans., 0.42.) 

103. Results by Special Engines. The Uniflow engine has 
unidirectional flow of steam, which is supposed to reduce con¬ 
densation. There is no exhaust valve, the long piston uncovering 
slots in the cylinder wall when near the end of its stroke. Fig. 34 



shows an engine with Corliss valves. Poppet valves are com¬ 
monly used. The engine is not well adapted for running non¬ 
condensing on account of the excessive compression which then 
results. A simple unjacketed engine of this type has given the 
remarkable steam rate of 9.06 lb. with 155 lb. pressure and 667° 
steam temperature (implying considerable superheat). 

The binary vapor engine uses two fluids. Steam is used 
in an ordinary cylinder. It then passes to a condenser, in which 
the cooling fluid is not water, but a volatile liquid (ether or S0 2 ). 
This “ binary fluid,” evaporated in the steam condenser, is 
used in a second cylinder and then discharged to a surface con¬ 
denser employing water as a cooling agent. The maximum 
efficiency is determined by the cooling water temperature and 
is ideally somewhat higher than that attainable by steam alone 
with present vacuum apparatus. Actual tests have given a heat 
rate around 167 B.t.u. per Ihp.-min. 

Regenerative engines have occasionally been built. Fig. 
35 illustrates the action in four stages (quadruple expansion). 
























































130 


Thermodynamics, Abridged 


The path cd represents adiabatic expansion in the first cylinder. 
The steam then passes to a reheater (Art. 100) on its way to the 
second cylinder. A portion of the steam is drawn off from 



Fig. 35. Quadruple Expansion Regenerative Eng'ne. 


this reheater and employed to heat boiler feed water. We may 
regard the heat withdrawn as the area under de, and the heat 
transferred to the feed water as the (ideally equal) area under ga. 
Similarly after expansion in the second cylinder, the heat under 
hf becomes the heat under kg: and after a third expansion the 
heat under Ij becomes the heat under pk. With an infinite 
number of stages, the equivalent expansion would be cq, the 
heat under which would equal the heat under pb. This is a 
regenerative cycle, like that in Art. 62, and the only heat 
chargeable is that under be. The efficiency is ideally (7&— T v ) / Tb, 
that of the Carnot cycle. Actual tests gave heat rates as low as 
169 B.t.u. per Ihp.-min. 

Prob. 200. A Uniflow engine has 2 per cent, clearance and 
opens its exhaust ports at 95 per cent, of the stroke. If the 
compression curve is pv = const., find the pressure at the end 
of the compression, the exhaust pressure being 1 lb. What is 
it if the exhaust pressure is 15 lb.? (Ans., 48| lb., 727^ lb.) 
















Efficiency and Power of Steam Engines 


131 


Prob. 201. A Uniflow engine uses 9.5 lb. of steam per Ihp.-hr. 
at 150 lb. pressure, superheated 300°. The back pressure is 
0.505 lb. Find the heat rate. (Ans., 206 B.t.u. per Ihp.-min.) 
Is any other value as low as this for a simple engine given in 
this chapter? 

Prob. 202. What is the ideal Carnot efficiency for steam at 
200 lb. initial pressure (dry) at 1 lb. back pressure? (Ans., 
0.332.) If by using a binary vapor the rejection temperature is 
reduced to 75° F., what is then the Carnot efficiency? (Ans., 
0.363.) If the relative efficiency as compared with the Carnot 
cycle is 0.45, what is then the heat rate? (Ans., 258 B.t.u. 
per Ihp.-min.) 

Prob. 203. Prove, in Fig. 35, that ybr = qcs. Prove that the 
efficiency is that of the Carnot cycle. 

Power or Capacity 

104. Mean Effective Pressure, Simple Engine, Saturated 

Steam. Methods of finding y m for ideal cycles have been given 
in Arts. 80, 81, 84, 85, 87. The usual method is based on the 
diagram ABCDF, Fig. 28, the curve BC being assumed to be, 
VbV b = ycVc • Then 

2JU(ft.lb.) ( Tr t T7 , V c T / \ • T7 
Vm = 144Fc = \P bVb + PeVslogey-- VdV d J-V c 

= + Ioge F^) - Vd ’ ( 57 ) 

in which y B = initial pressure, y D = back pressure, Vc/V B 
= ratio of expansion. The maximum overload capacity (with 
no cut-off) is 

AHDF - ABCDF _ P B - F D -y m f5g) 

ABCDF ym { 

Values of y m from Equation (57) are multiplied by a diagram 
factor, /, (sometimes called the “ card factor ”) to give the 
probable mean effective pressure in the actual engine. Note that 
/ = abceg -5- ABCDF. Values of / range from 0.6 to 0.9, de- 





132 


Thermodynamics, Abridged 


pending chiefly on the valve gear. Jackets increase / by 0.05 
to 0.15. Its value is decreased by high ratios of expansion, 
speeds above 225 r.p.m., excessive clearance, and inadequate 
valve passages. 

Piston speeds are usually between 500 and 1000 ft. per min. 
The allowable r.p.m. depend chiefly on the valve gear. Long 
strokes favor low clearances. See Art. 7. 

Prob. 204. With an initial pressure of 115 lb., a back pressure 
of 2 lb. and a ratio of expansion of 4.5, find p m and ideal overload 
capacity. If / = 0.85, piston speed = 800, r.p.m. = 100, find 
the cylinder diameter and stroke, the engine being rated at 500 
Ihp. (Ans., p m = 62, ideal overload capacity = 82 per cent., 
diameter = 22.3 in., stroke = 48 in.) 

105. Superheated Steam. For superheated steam, the expan¬ 
sion curve BC, Fig. 28, is “ steeper ” than for saturated steam. 
Its equation is PbV s a = PcVc a , where a may have values about 
as follows: 


Ratio of Expansion 

Superheat, °F. 

200 | 

250 

300 1 

350 

400 

450 

500 

Values of Exponent a 

8.0 

1.05 

1.07 

1.09 

1.11 

1.13 

1.15 

1.18 

7.0 

1.06 

1.08 

1.10 

1.12 

1.14 

1.16 

1.19 

6.0 

1.07 

1.09 

1.11 

1.14 

1.16 

1.18 

1.21 

5.0 

1.08 

1.10 

1.12 

1.15 

1.17 

1.19 

1.22 

4.0 

1.09 

1.11 

1.13 

1.16 

1.18 

1.20 

1.23 

3.0 

1.10 

1.12 

1.15 

1.17 

1.20 

1.22 

1.25 


The mean effective pressure is now 

V B a (Vs\ a 1 

^Vb Tc -~i - Pb[ Vc ) ■ —! - Vr>- (59) 

This gives a lower value than Equation (57)—see Art. 98— but 
(as with jackets) the value of f is higher. With slight super¬ 
heat, use Equation (57) for p m but use high values of f in con¬ 
nection therewith. 
















Efficiency and Power of Steam Engines 


133 


Prob. 205. Under the conditions of Prob. 204, what is the 
value of p m for steam superheated 100°? (Ans., 62.) If the 
steam is superheated 300°, what is the probable value of a? 
Of p m ? (Ans., a = 1.125, p m = 59.) 

106. Compound Engine: Ratios. In the preliminary propor¬ 
tioning of a compound engine, it is assumed that the combined 
diagrams of the two stages will appear as ABCDF, Fig. 28, and 
that the steam exhausted from the high-pressure cylinder along 
JK enters the low pressure cylinders along KJ without fall of 
pressure or change of volume. The curve BC is assumed to be 
(with saturated steam) p B V B = pcVc . The ratio of cylinder 
volumes (cylinder ratio) is C = Vc/Vj = pjlpc • For cyl¬ 
inders of equal strokes (the usual case), C = AijAn, where Ai 
and A h are the cross-sectional areas of low and high pressure 
cylinders. But pc = PbI-v, where r = whole ratio of expansion. 
Then C = rpjjp B . The receiver pressure (receiver between 
the cylinders) is pj. The ratio of expansion in the high pressure 
cylinder is r h = Vj/V B = p B \pj . That in the low-pressure 
cylinder is V c /Vj = C. 

There are five methods of establishing a value for p Jt 

(1) It may be assumed. 

(2) A value may be assumed for C. Then pj = Cp B /r. 

(3) The temperature ranges may be equalized in the two 
stages. Then t B — tj = tj — t D , tj = ( t B + <d)/ 2. Values of 
t B and t D are those from the steam table, corresponding with the 
pressures p B and p B - The pressure pj is then the tabular 
pressure for tj. 

(4) Maximum total piston pressures may be equalized, i.e., 

Pb - Pj= C( P j - p D ) 

Pj = Vb — C(pj — p D ) 

= PB - (Pj - Vd) 

p B 

= Vb 2 (.PB + *Pj ~ r Vo)- 


( 60 ) 


134 


Thermodynamics, Abridged 


This is the important method for marine practice. 

(5) The two stages may develop equal power. Then ABJK 

= KJCDF, or 

PbVb loge ^ 5 = PjVj + PjVj logel^ - PdVd 
Vj y j 

log. p B - loge Pj = 1 + log. C - r p 

Pb 

= 1 + log. Pj - loge Pb + loge T ~ T 

V B 

log. Pj = |(log. r + - l)- log. jn (61) 

Prob. 206. Given initial and back pressures of 120 and 2 lb., 
with r = 25, find pj under each of the following conditions: 

(a) (7=4. (Ans., 19.2.) 

(b) Temperature ranges equalized. (Ans., 22.) 

(c) Maximum total pressures equalized. (Ans., 22.7.) 

(d) Power division equalized. (Ans., 17.8.) 


107. Approximate Dimensions. After having determined pj 
by one of the foregoing methods, the mean effective pressures 
in the high and low pressure cylinders are 


Pmh 

Pml = 


ABJK 

Vj 

KJCDF 

V D 


Pb, 

= ^ (1 + loge C) — Pd 


(62) 


The output of the engine is 


hp. = 


fS 

33 000 


(:Pmh Ah + Pml A I ) 


fSA, 
33 000 



from which Ai is determined. The value of / is about the same 
as that in a simple engine using a ratio of expansion equal to Vr. 
The quantity ( pmh/C + p m i) is called the total mean effective 

PRESSURE REFERRED TO THE LOW-PRESSURE CYLINDER. Its 

value is the same as that given by Equation (57), applied to the 
whole diagram ABCDF. Then 







Efficiency and Power of Steam Engines 


135 


^ = 33Wo{7 (1 + lo ^ r) -M- (63o) 

It thus appears that the size of the high pressure cylinder 

IS WITHOUT INFLUENCE ON THE TOTAL POWER OF THE ENGINE. 

Prob. 207. The initial, back and receiver pressures are 120,2 
and 17.8. The ratio of expansion is 25. The engine develops 
500 Ihp. at 800 ft. piston speed and 100 r.p.m. Find cylinder 
diameters and strokes. Use / = 0.9. (Ans., 20.8 and 40 in. 
by 48 in.) 

Prob. 208. In Prob. 207, what are the mean effective pressures 
in each of the two cylinders? (Ans., 33.9 and 9.11.) Show that 
the two cylinders do develop equal power. 

Prob. 209. If in the foregoing two low pressure cylinders are 
used, state dimensions. (Ans., all strokes 48 in.: diameters, 
high pressure, 20.8 in.; two low pressure, 28.5 in.) 

Prob. 210. Given p m h = 35, p m i = 9, C = 4, in a compound 
engine. What proportion of the total power is developed in each 
cylinder? (Ans., high pressure 49.2 per cent., low pressure 50.8 
per cent.) 

108. Cylinder Ratio and Drop. The value of C has not been 
thoroughly standardized. It should increase with r. High 
values increase the initial steam pressure in the second stage. 
The margin of overload capacity, obtainable by admitting 
reduced pressure steam from the boilers to the low-pressure 
cylinder, is thus decreased. Values up to 6 or 7 have given 
highly efficient engines. The ideal diagram of Fig. 28 shows 
terminal drop, CD, in the low pressure cylinder but none in 
the high pressure cylinder. Drop (high pressure drop) is often 
provided for. This reduces the value of / about 0.05. In 
general practice, the mean effective pressure referred to the low 
pressure cylinder ( pmh/C + p m i) is kept around 1.2 + 0.12p B . 

Prob. 211. Following the rule in Art. 108, what would have 
been the diameter of the low pressure cylinder for 500 Ihp. at 
800 ft. piston speed with / = 0.9? (Ans., 43 in.) 

10 



136 


Thermodynamics, Abridged 


109. Compounds with Superheat. For superheat below 150°, 
the foregoing method may be used, increasing / as when jackets 
are employed. With high superheat, take the value of a in 
Art. 105 which would be suitable for a ratio of expansion equal 
to Vr in a simple engine. Then, following Equation (59) 

_ Vj^ _ Vj ■ 

Vml ~ C{a - 1) C a (a - 1) Pd 



In this case pj = pB(C/r) a and r h = r/C. If the work of the 
two cylinders is to be equally divided, it will be found that 

»+« , -'+ i }]""“• 

The horse power is 

_ fSAi (Pmh \ 

hp ’ 33 000\ C + P ml )’ 

where the quantity in the parenthesis has the value given by 
Equation (59). 

Prob. 212. For r = 25 at 300° superheat, what value of a 
should be used? (Ans., 1.12.) If p B = 120, p D = 2 , Pj — 20, 
find C. (Ans., 5.05.) Find the value of C for equality of work. 
(Ans., 5.2.) What is then the value of p 7 ? (Ans., 20.6.) 
Under the latter condition, find Th, p m h, Pmi • (Ans., = 4.8, 
p m h = 40^, pmi = 7.85.) What is the value of the total mean 
effective pressure referred to the low pressure cylinder, as deter¬ 
mined by Equation (59)? (Ans., 15.7.) 

110. Triple Engines: General Case of Multiple Expansion. 
For any multiple expansion engine using saturated steam (or 
steam superheated less than about 150° F.), the size of the low 
pressure cylinder is given,by Equation (63a). The sizes of the 
other cylinders, for zero drop, are then determined from assumed 







Efficiency and Power of Steam Engines 


137 


values of cylinder ratios or receiver pressures. Their sizes 
influence efficiency but not power. 

With high superheat, similarly, following Equation (59), 


i _ fSAi f p B a _ Vb 

P ' 33 000 \ r ' a - 1 r\a - 1) P ° 


(65) 


If b is the number of expansion stages, take a from Art. 105 for 
a ratio of expansion of Vr. 

Prob. 213. The battleship Oklahoma uses triple engines at 
280 lb. initial absolute pressure. Assuming 1 lb. back pressure, 
and 10 000 Ihp. capacity at cruising speed when r — 30, with 
/ = 0.9, what would be the diameter of each of the two low 
pressure cylinders at 1000 ft. piston speed, using saturated steam? 
(Ans., 76 in.) 

Prob. 214. The cylinder ratios in the Oklahoma are 1 : 2.84 : 
9.93 (high — intermediate — low). Find the diameters of high 
and intermediate cylinders under the conditions of Prob. 213. 
(Ans., 34 and 64 in.) 


Prob. 215. Under the conditions of Prob. 213, but with 
steam superheated 300°, what is the probable value of a? (Ans., 
1.15.) What is then the diameter of the low pressure cylinder? 
(Ans., 84^ in.) 





CHAPTER VI 


VAPOR REFRIGERATION 

111. Ideal Cycle. The ideal cycle in vapor refrigeration is a 
Rankine cycle worked counter-clockwise (Fig. 36). Assume 
the fluid to be at low temperature and pressure, somewhere on 
the line eb, but nearer e than b, i.e., in the condition of a very 
wet vapor. It is contained in coils in the room or tank to be 
cooled, the contents of such room or tank being warmer than 
the fluid. Heat will flow from such contents to the fluid. This 
will not increase the temperature of the fluid, but will vaporize 
whatever liquid it contains. 

When the vaporization is nearly (but not quite) complete, 
allow the fluid to flow out to a compressor. Here it is com¬ 
pressed adiabatically until its temperature is above that of an 
available supply of cooling water. The operation is indicated 
by the path be. The pressure will rise, as well as the temperature. 
The fluid now passes to the condenser, where it is subjected to 
the cooling effect of water in the coils. This water condenses 
the vapor (path cd) but in general cannot cool it much. The 
final operation necessary is a reduction in temperature of the 
resulting liquid. In air refrigerating machines (Arts. 57, 60) 
the final operation took place in an expanding cylinder. 
With air, mere expansion from a high pressure to a lower pressure 
would not produce cooling, unless at the same time work is 
done. With a vapor, on the other hand, expansion alone will 
reduce the temperature. Hence the final operation, which in 
Fig. 36 reduces the temperature of the liquid from td to t e , is 
merely a reduction of pressure accomplished by allowing the 
fluid to flow through an expansion valve. A high pressure 
exists from the compressor outlet to the expansion valve inlet. 
A low pressure exists from the expansion valve outlet through 
the cold room to the compressor inlet. 

138 


Vapor Refrigeration 139 

112. Choice of Fluid. It is desirable that the high pressure 
(necessarily the pressure corresponding with a normal cooling 



Entropy 


Fig. 36. Vapor Refrigeration with Ammonia. 

water temperature of say 80° F.) be not too high. It is also 
desirable that the low pressure (that corresponding with the 



















































































140 


Thermodynamics, Abridged 


lowest temperature of the fluid, which for ice-making and general 
service is about 0° F.), be not too low. The latter pressure 
should preferably be above the pressure of the atmosphere. 
If it is lower than this, any leakage will be inward. Leakage 
cannot then be readily detected, and the fluid will be diluted if 
leakage occurs. Fig. 20 shows that from the first standpoint the 
best fluids are ether, ethyl chloride, sulphur dioxide and ammonia. 
Carbon dioxide, at 80° F., gives a pressure of nearly 1000 lb. per 
sq. in. From the second standpoint, ether, ethyl chloride and 
S0 2 are all unsatisfactory, giving pressures below that of the 
atmosphere at 0° F. The fluids mostly used are S0 2 , ammonia 
and C0 2 . The last is most popular in marine service, since it is 
innoxious. In the following discussion, reference is made 
particularly to ammonia. This is for two reasons. Its proper¬ 
ties are fairly well known. It permits greater flexibility of 
cycle, because there is in no case any approach to its critical 
temperature (Art. 65). 

113. Action in the Expansion Valve. A new type of vapor 
process is here encountered, which will be more fully discussed 
later (Chapter VIII). The liquid at d, Fig. 36, is converted to 
fluid at the pressure p e without change of heat content. 
Since hd > h e , the fluid cannot be liquid at the new pressure. 
It can only be a wet vapor. Its state is /, defined by heat at 
/ = heat at d, or hf + (xL)f = hd. We must regard the process 
as being represented by def. It is not an adiabatic process. 
The refrigeration that can be performed by the fluid is then not 
the area under eb, but only the area under fb. 


Properties of Saturated Ammonia Vapor 


t 

Temperature, 
Deg. F. 

P 

Pressure, 
Lb. per 
Sq. In. 

V 

Volume, 
Cu. Ft. 
per Lb. 

h 

L 

H 

r 

n w 

n e 

n* 

0 

29.95 

9.190 

-33.7 

572.2 

538.5 

521.4 

-0.0709 

1.2450 

1.1741 

70 

129.20 

2.296 

42.1 

512.8 

555.0 

458.5 

0.0813 

0.9684 

1.0497 

90 

181.80 

1.650 

65.3 

493.5 

558.9 

438.9 

0.1238 

0.8981 

1.0219 

95| 

199.00 

1.512 

71.9 

488.0 

559.9 

433.3 

0.1356 

0.8790 

1.0146 


(From Goodenough’s “ Properties of Steam and Ammonia”) 














Vapor Refrigeration 


141 


Prob. 216. Liquid ammonia at 70° expands to 0°. What is 
the dryness of the resulting vapor? (Ans., 0.132.) What is 
its entropy? (Ans., 0.0936.) 

Prob. 217. This vapor in Prob. 216 is 0.87 dry when it leaves 
the coal room. How much refrigeration has been done per lb. 
of fluid? (Ans., 422 B.t.u.) 


114. Ideal Power and Efficiency. In symbols, per lb. of fluid, 
and in B.t.u., 

Refrigeration done = heat under fb = L/b(xb — xf) 

Heat rejected at condenser = heat under cd = L c d(x c ) 

Work expended = heat at c — heat at b — (h-\- xL) c — (h + xL\ 


Efficiency = 


Refrigeration 

Work 


Lfb(x b — x f ) 

( h + xL) c — (h+ xL) b ' 


( 66 ) 


The last may exceed unity. The value of x c is found from 


(n w + xn e ) c = (n w + xn e ) b . 


Efficiencies are much higher than with “ dense air ” machines 
(Art. 60). The mean effective pressure is 


work X 5.403 
maximum volume 


see Equation (67). 


Prob. 218. The fluid is ammonia. Temperature limits are 
0° and 70°. The fluid is 0.87 dry at the beginning of com¬ 
pression. Find the dryness at the end of compression. (Ans., 
0.96.) 

Prob. 219. In Prob. 218, how much heat is removed by the 
circulating water at the condenser? (Ans., 494 B.t.u. per lb. 
of fluid.) 

Prob. 220. In Prob. 218, how much work is expended? 
(Ans., 70.8 B.t.u. per lb. of fluid.) What is the ideal efficiency? 
(Ans., 5.98.) 

115. Capacity, Power and Size. If T = tonnage (Art. 59), 
the necessary refrigeration is 200 T B.t.u. per min. Then the 
weight of fluid to be circulated per minute is w = 200 T -f- re¬ 
frigeration per lb. and the ideal hp. is (w/42.42) {(h + xL) c 





142 


Thermodynamics, Abridged 


— (h + xL)b}. The heat rejected at the condenser is wL cd x c 
B.t.u. per min. A close approximation to the compressor dis¬ 
placement may be made as follows: 

The volume of fluid present at the beginning of compression 
is Vb in the pv diagram. (Lettering on the pv diagram corre¬ 
sponds with that on the Tn diagram.) This is substantially 
( xv\ . Assume clearance = c and assume that the clearance 
expansion curve is p g v g = ph$h- Then Vh = cD(p g /ph), where 
D = displacement of compressor. The volume of fresh fluid 
drawn in and compressed is Vb — v h = (1 + c)D — cD(p g /ph) 
= D{1 + c(l — (pg/ph))}- The weight, w, is this quantity 
divided by ( xv)b , the specific volume of the wet vapor. The 
weight being known, the displacement may be found. 

Thus, suppose clearance to be 4 per cent., and p g /ph — 5 . 
The volume compressed is B{1 -f- 0.04(1 — 5)} = 0.84 D. Note 
that 0.84 is the volumetric efficiency (Art. 42), suction fric¬ 
tion being ignored. Suppose the vapor to be 0.90 dry at the 
beginning of compression, when its temperature is 0°. Then 
(xv)b = 0.90 X 9.19 = 8.271. Suppose 100 lb. of fluid to be 
circulated per min. Then the volume to be circulated is 827.1 
cu. ft. = 0.84 D whence Z) = 983 cu. ft. per min. 

The equation given for mean effective pressure in Art. 114 
may now be written 


Pm = 


5.403{(ft + xL) c - (ft+ xL)b] 
C xv) b 


(67) 


Prob. 221. In Probs. 218-220, the capacity of the machine 
is 100 tons. Find the weight of fluid per min. and the ideal hp. 
(Ans., 47.4 lb., 79 hp.) 

Prob. 222. In the foregoing, how much heat must be carried 
away by the cooling water? (Ans., 23 400 B.t.u. per min.) 

Prob. 223. In Prob. 221, the clearance is 5 per cent., the 
piston speed 240 ft. per min., the compressor double-acting at 
60 r.p.m. Find the diameter and stroke of the compressor 
cylinder. (Ans., 18.6 and 24 in.) What is the mean effective 
pressure? (Ans., 48 lb. per sq. in.) 



Vapor Refrigeration 


143 


Prob. 224. In Prob. 221, what is the ideal ice-melting effect 
per Ihp.-hr.? See Art. 59. (Ans., 106.) 

Actual Performance. In the foregoing illustrative case, the 
condenser temperature (70°) is lower than may in general be 
safely assumed. Consequently the ideal efficiency is higher 
than is to be expected in ordinary plants. A narrow temperature 
range leads to efficiency in refrigeration. An ideal ice-melting 
effect (Art. 59) of about 80 lb. per Ihp.-hr. is attainable between 
limits of 0° and 90°. About 60 lb. would be considered a fairly 
good result if actually attained in ordinary practice, with these 
limits. There is thus implied a relative efficiency of 60/80 
= 0.75. This is higher than the value of relative efficiency 
attainable with steam engines, mainly because the temperatures 
involved are closer to those of the surrounding atmosphere. 
The horse power of the engine which drives the compressor is 
greater than the horse power here computed, the ratio of the 
two being the mechanical efficiency. Under the 0°-90° 
limits, one horse power in the ammonia compressor cylinder 
should give a trifle less than one ton of capacity. Under un¬ 
favorable conditions, as much as 2 hp. per ton may be required 
at the steam engine cylinder. (Capacity is ice-melting effect, 
not ice-making capacity.) 

Piston speeds range from 125 to 600 ft. per min., being low 
in small machines. Mean effective pressures vary widely (30 
to 90 lb.) with the assigned temperature limits. 

In purchasing a compressor, the stated tonnage should be 
checked against the displacement, for the temperature limits to 
be employed. It is seldom safe to assume a condenser tempera¬ 
ture below 90° for year-round shore service, but 75° may be 
realized on ships in northern latitudes. A low temperature of 0° 
will just about answer for ice-making with direct expansion 
(ammonia circulated in coils in the freezing tank). A slightly 
lower temperature still will accelerate the freezing. When the 
brine system is used (ammonia cools the brine and brine does 
the refrigeration) the possibly offensive fluid is kept away from 
substances it might injure, but efficiency is sacrificed. The low- 


Specific Heat 


144 


Thermodynamics, Abridged 


est ammonia temperature must be below the lowest brine 
temperature, and the latter must be below the desired cold-room 



Specific Gravity (Water-1) 


Fig. 37. Properties of Brines Plotted^Against Specific Gravity. 

temperature. Typical values for the three temperatures are 
0°, 15°, 25°. If brine is used, the ammonia will work between 








































































































































































Vapor Refrigeration 


145 


0° and 90°. With direct expansion, it might have worked be¬ 
tween 15° and 90°. The latter range gives a more efficient cycle. 
Direct expansion is greatly to be preferred on economical grounds 
for ice-making or very low temperature work. Brine circulation 
gives opportunity for storage of “ cold ” and no harm is done 
if the compressor has to be shut down for a short period. 

In indirect refrigeration, air is used instead of brine. Air 
is blown over coils carrying the refrigerating fluid, and the 
chilled air then passes to the room to be cooled. 

Fig. 37 gives the properties of the two types of brine used. 
Calcium chloride permits of lower temperatures than salt brine, 
and is less corrosive. 

Prob. 225. In Prob. 224, the machine as actually installed 
developed 98 tons capacity and the compressor cylinder indi¬ 
cated 107 hp. while the Ihp. of the steam engine driving the 
compressor was 133.33. State the relative efficiency and the 
mechanical efficiency. (Ans., 0.72, 0.80.) If the actual indi¬ 
cated thermal efficiency of the steam engine was 0.10, what was 
the efficiency from steam at the throttle to refrigeration? (Ans., 
0.38.) 

Prob. 226. What is the Carnot efficiency of refrigeration 
between 0° and 90° F.? (Ans., 5.1.) Between 15° and 90°? 
(Ans. 6.3.) 

Prob. 227. The machine in Prob. 221 is used for indirect (air- 
blast) refrigeration. The air is cooled at constant pressure from 
70° to 10°. Ignoring radiation loss, what weight of air is circu¬ 
lated per min.? (Ans., 1400 lb.) 

Prob. 228. A calcium chloride solution has a freezing point 
of — 20°. What is its strength? (Ans., 24.4 per cent.) What 
is its specific gravity? (Ans., 1.217.) Its specific heat at 5° F.? 
(Ans., 0.688.) If with a 100 ton machine its temperature rises 
from — 15° to 25° F., what weight of solution must be pumped 
per min.? (Ans., 726 lb.) 

116. Tonnage Rating. The 100 ton machine of Prob. 223 has 
a displacement of 454 cu. ft. per min., or 4.54 X 1728 = 7840 


146 


Thermodynamics, Abridged 


cu. in. per ton per min. This figure (the specific displacement) 
depends on the clearance, the limiting temperatures and the 
dryness at the beginning of compression. There is a tendency 
to standardize 0° and 95|° as limits for rating. Under these 
conditions a specific displacement proposed is 5.52 cu. ft. or 
9570 cu. in. 

Prob. 229. In Art. 116, if clearance is 5 per cent., and temper¬ 
ature limits are 0° and 95^°, what value of Xb would give a specific 
displacement of 9570? (Ans., 0.94.) 

117. Dry Compression. In Prob. 229, the vapor is nearly dry 
at the beginning of compression. Inspection of Fig. 36 will 
show that it is bound to become somewhat superheated during 
adiabatic compression to 95|°. Such operation is described as 
“dry compression,” in distinction to “wet compression” hereto¬ 
fore described. Wet compression leads to the more efficient 
cycle, especially if Xb is so adjusted that x c = 1.0. This is a 
condition difficult to maintain, and it is obviously not the 
condition which gives maximum refrigeration per lb. of fluid. 
Such maximum refrigeration is obtained (without allowing the 
fluid temperature to rise in the cold room) when Xb = 1.0. In 
order to maintain wet compression under this condition, some¬ 
thing must be done to withdraw heat during compression. This 
makes the path depart from the adiabatic, swinging it to the 
left as it proceeds upward. Jackets can be used to accomplish 
this result, as in air compressors. Multi-stage operation is 
occasionally practiced. The injection of cold oil or of liquid 
ammonia into the cylinder is frequently adopted. In any case, 
the rate of flow of fluid should be increased as soon as the com¬ 
pressor discharge becomes noticeably warm. This indicates 
(with normal values of p c ) the existence of some superheat. 
Increasing the flow of fluid will decrease Xb and reduce or eliminate 
the superheat. 

Prob. 230. With 0° and 90° limits, find xt when x c = 1.0. 
(Ans., 0.88.) 


Vapor Refrigeration 


147 


Prob. 231. While steadily giving 100 tons of refrigeration, 
the machine in Prob. 217 is speeded up so that the weight of 
fluid circulated per min. is increased 10 per cent. What change 
occurs in xrf (Ans., it is reduced from 0.87 to 0.80.) 

118. Ideal Cycle with Superheat. In ideal dry compression 
without heat removal, we have such a cycle as ejkld, Fig. 36. 
The point k is not located to scale. The line Ik is a line of 
constant pressure. We now have, following Art. 114, 


Refrigeration done = Lfjil — Xf), 

Heat rejected at condenser = H k ' — hd, 
Work expended = Hk — Hj, 


Efficiency Hk _ H % 1 

Mean effective pressure = p m = 
Volumetric efficiency as in Art. 115. 


5.403(17*' - Hi) 


The value of Xf is found as before, in Art. 113. That of Hk is 
best found from a total heat-entropy diagram, Fig. 38. The 
entropy and pressure at k are known. The symbol Hj means 
the total heat of dry vapor at j. 

Thus take limits of 0° and 70°. From Prob. 216, xj = 0.132. 
Then 

Refrigeration done = 572.2 (1 — 0.132) = 497, 

Pressure at k = 129.2, 

Entropy at k = (n s ) 0 o = 1.1741, 

H k r from Fig. 38 = 630, 

Heat rejected = 630 — 42.1 = 587.9, 

Work = 630 - 538.5 = 91.5, 

Efficiency = 497 -f* 91.5 ■= 5.45, 
p m = (5.403 X 91.5) -s- 9.19 = 54. 

The efficiency is less than in Prob. 220 and the mean effective 
pressure is necessarily greater. For 100 tons capacity, 40.25 lb. 
of fluid must be circulated per min. This is of course less than 
in Prob. 221. The volume of fluid per min. is 40.25 X 9.19 
= 370 cu. ft., and with volumetric efficiency = 0.834 as in Prob. 




148 


Thermodynamics, Abridged 



o2£ 3A °qV W>1 


0.95 1.00 ‘ 1.05 1.10 1.15 1.2 

Entropy 

Fig. 38. Total Heat-Entropy Diagram for Superheated Ammonia. (Plotted from Goodenough’s “Properties of Steam 

and Ammonia.”) 

















































































































































































































Vapor Refrigeration 


149 


223, the specific displacement is (3.7 X 1728) 4- 0.834 = 7700 
cu. in., a value not notably different from that realized in wet 
compression. 

Prob. 232. Find the ideal efficiency with dry compression 
from 0° to 90°. (Ans., 4.12.) 

119. Carbon Dioxide Machine. The temperature-entropy 
diagram, Fig. 39, is plotted with lines of constant pressure and 
constant superheat in the superheat region. Dry compression 



- 0.05 0 0.05 0.10 0.15 020 025 0.30 0.35 

Entropy 

Fig. 39. Temperature-Entropy Diagram for Carbon Dioxide. 


is common, but leads to low efficiency. Adiabatic compression 
from a dry state at 0° to the pressure of 900 lb. (corresponding 
roughly with a saturation temperature of 75°) would put the 
final temperature far above 100°. In the tropics, the vapor 
could not then be condensed. We must in practice in any 
latitude either compress from an initially wet condition, or 
remove heat during compression, or be content with com¬ 
paratively low efficiency. 

































































































































150 


Thermodynamics, Abridged 

Properties of Saturated Carbon Dioxide Vapor 


t, 

Temperature, 

Deg. F. 

P> 

Pressure, 

Lb. per Sq. In. 

V, 

Volume, 

Cu. Ft. per Lb. 

Vw, 

Volume of Liquid, 

Cu. Ft. per Lb. 

h 

L 

H 

r 

n w 

n e 

n» 

0 

75 

313.5 

902.0 

0.2661 

0.0708 

0.016298 

0.022232 

-15.5 

29.4 

i 

114.3 

50.8 

98.8 

80.2 

99.81 

42.69 

-0.03278 

0.05718 

0.2494 

0.09495 

0.21662 

0.15213 


In the first case, the method of analysis is precisely as in Arts. 
113, 114. In calculating mean effective pressure and displace¬ 
ment, however, it should be noted that the values of v and v w 
are too nearly the same to permit of the approximation hereto¬ 
fore used (Arts. 74, 115) for the specific volume at the beginning 
of compression. That volume is for vapor at 0° F., 0.90 dry, 
for example, not 0.90 X 0.2661, but 0.90 (0.2661 — 0.016298) 
+ 0.016298 = 0.2418. 

Prob. 233. With C0 2 between 0° and 75°, the vapor is 
exactly dry at the end of adiabatic compression. What was its 
dryness at the beginning of compression? (Ans., 0.743.) 

Prob. 234. In Prob. 233, what was the dryness at the expan¬ 
sion valve outlet? (Ans., 0.39.) Compute the refrigeration per 
lb. of fluid. (Ans., 40 B.t.u.) How much heat is rejected at 
the condenser? (Ans., 51 B.t.u. per lb.) How much work is 
done? (Ans., 10.5 B.t.u. per lb.) What is the ideal efficiency? 
(Ans., 3.83.) What is the specific volume at the beginning of 
compression? (Ans., 0.202.) What is the volumetric efficiency 
if clearance is 4 per cent.? (Ans., 0.9248.) What is the dis¬ 
placement per lb. of fluid? (Ans., 0.218 cu. ft.) What is the 
specific displacement — cu. in. per min. per ton — for the con¬ 
ditions assigned? (Ans., 1870.) 

120. Carbon Dioxide Cycle with Heat Removal. Suppose 
heat to be removed during compression at such rate as to make 
the vapor dry at both the beginning and end of compression. 
The cycle is represented diagramatically (not to scale) by ejld, 















Vapor Refrigeration 


151 


Fig. 36. Assume the compression path jl to be a polytropic. 
Then pivf = pjVj a , and for the 0°-75° limits, the exponent is 

a _ log pi — l og Vj = log 902 - log 313.5 
log Vj — log vi ~ log 0.2661 — log 0.0708 

2.955 - 2.496 
- 0.5744 + 1.15 ~ °' 798- 

Then the external work done under this curve jl is 

w _ Pfli ~ PM = (902 X 0.0708) - (313.5 X 0.2661) 
jl a - 1 -0.202 

144 

X 778 = 17.9 B.t.u. 

The external work of the whole cycle is 
2PF = W ed + W dl + Wij - W je 

_ (pVw)d Pl(v — v w )i Pj{v - x w )j 

5.403 ^ 5.403 A/,y 5.403 

_ (pv)i pj(v - v w )j 902 X 0.0708 

5.403 5.403 “ 5.403 

313.5(0.2661 - 0.0163) 
+ 17,9 5.403 - 

= 11.8 + 17.9 - 14.5 = 15.2 B.t.u. 

The refrigeration done is Lj( 1 — x/). The heat rejection at 
the condenser is Li. The specific volume at the beginning of 
compression is the tabular volume Vj. Since Si/ = 2 W, 

Hi + Hij - Hj = 15.2 = 80.2 + H tj - 98.8 
Hij = 33.8 B.t.u. per lb. of fluid. 

This is the quantity of heat to be removed during compression. 

Prob. 235. In the foregoing, find refrigeration and heat rejec¬ 
tion per lb. of fluid, and the ideal efficiency. (Ans., 69.6, 50.8, 
B.t.u.: efficiency = 4.59.) 

Prob. 236. The machine in the foregoing is of 100 ton 
capacity, double-acting, at 240 ft. piston speed and 5 per cent. 
11 












152 


Thermodynamics, Abridged 


clearance. Find cylinder diameter, ideal hp. and ideal ice¬ 
melting effect per Ihp.-hr. (Ans., 8 in., 103 hp., 81 lb.) 

Prob. 237. In Probs. 235-236, 33.8 B.t.u. must be removed 
per lb. of fluid during compression. If this is accomplished by 
injecting liquid at 0° into the cylinder, what weight of fluid 
must be thus supplied per min., the injected liquid being com¬ 
pletely vaporized at 75°? (Ans., 102 lb.) 

121. Comments. This last type of cycle requires less power 
for a given amount of refrigeration, and is consequently more 
efficient, than the cycle with adiabatic compression: but it 
requires jacketing or liquid injection. Carbon dioxide gives 
lower thermal efficiencies than ammonia, but much higher 
efficiencies than air. On account of its low specific volume (high 
density) the displacement is less in proportion to refrigerating 
capacity than is the case with ammonia. The very high pressures 
reached limit the application of carbon dioxide to cylinders of 
small diameter. Hence large machines rarely use this fluid. 
Leakage is more probable than with ammonia, and is not easy 
to detect. The fluid is usually cheaper than ammonia. For 
extremely low temperatures, carbon dioxide can be used with¬ 
out maintaining a vacuum between the compressor and the 
expansion valve. 

122. Sulphur Dioxide. This fluid is occasionally employed in 
marine service. The following values are given by Zeuner: 


Properties of Saturated Sulphur Dioxide 


1 t. Temperature 

| Fahrenheit 

p, Pressure, Lb. 
per Sq. In. 

v. Volume. Cu. 
Ft. per Lb. 

h 

L 

H 

r 

M'w 

n e 

n» 

-4 

5 

77 

86 

95 

9.272 

11.756 

56.386 

66.359 

77.630 

8.051 

6.486 

1.445 

1.220 

1.036 

- 11.216 
- 8.449 
14.582 
17.572 
20.588 

171 

169.745 

148.775 

144.787 

140.495 

159.784 

161.296 

163.357 

162.359 

161.083 

157.111 

155.563 

133.718 

129.827 

125.656 

- 0.0237 

- 0.0177 

0.0284 

0.0339 

0.0394 

0.3755 0.3518 ' 
0.3655 0.3478 
0.2773 0.3057 
0.2655 0.2994 
0.2534 ' 0.2928 


Prob. 238. What property, in the above table, varies in an 
unusual way? 















CHAPTER VII 


HEAT TRANSFER APPLIANCES 

123. Mixtures. Mixtures of simple substances have been 
considered in Art. 3. For these, or for vapors, ignoring radia¬ 
tion, 

loss of heat = gain of heat, 

Wa(Qa ~ Qa ) = W b (Q b ' ~ Q b ) , 

where w = weight in lb., Q = heat content in B.t.u., and primes 
denote the high temperature conditions. In actual mixing, 
t b = t a : i.e., the final temperatures of the two fluids are the 
same. For simple substances, w(Q' — Q ) is best written, 
ws(t' — t), where s = specific heat. For vapors, 

Q = h at the liquid condition, 

Q = h + xL when wet, 

Q = II when dry, 

Q = H + k(t B up — 4at) when.superheated. 

Prob. 239. Ammonia at 199 lb. pressure, superheated 220°, 
is cooled, condensed and further cooled until its temperature 
is 70° F. See Fig. 38. How much heat is emitted per lb.? 
(Ans., 598.9 B.t.u.) 

Prob. 240. An injector is supplied with dry steam at 200 lb. 
pressure and water at 90° F. The two fluids mix and the 
resulting temperature is 170.06° F. What weight of water was 
mixed with 1 lb. of steam? (Ans., 13.3 lb.)* 

Prob. 241. In a jet condenser, 75.07 lb. of water are employed 
per lb. of steam. Water enters at 80° F. Steam enters at 2 lb. 
pressure, 0.70 dry. What is the temperature of the discharged 
mixture? (Ans., 90° F.) 

124. Temperatures in Surface Transfer. Transfers through 
surfaces (like the metal wall of a condenser tube) are more 


* See Art. 134. 


153 


154 


Thermodynamics, Abridged 



complicated. There are four temperatures to be considered; 
the inlet and outlet temperatures of both the heat-emitting and 
the heat-receiving fluids. The rate of transfer depends not 

only on the values of these 
four temperatures, but also, 
so to speak, on their ar¬ 
rangement. Take the case 
of an economizer—a group 
of tubes containing feed 
water, placed in the path of 
the combustion gases from 
a boiler to its stack. The wa¬ 
ter moves vertically through 
the tubes, but it also moves 
(considering the whole sys¬ 
tem of tubes) in a direction 
which may be either toward 
the boiler or away from it. 

The two cases thus sug¬ 
gested are represented dia- 
grammatically in Fig. 40. 
In PARALLEL FLOW, the two 
fluids move in the same gen¬ 
eral direction. In counter 
flow, they move in opposite 
directions. In many devices, the direction of flow is uncertain 
or indeterminate. Such is the case with a steam radiator, for 
instance. 

The rate of heat flow is generally proportional to the tempera¬ 
ture difference, that is, to the difference between the temperatures 
of hot and cold fluids. The total heat transfer is proportional 
to the mean temperature difference, t m .. This is the average 
ordinate or vertical distance between the two curves of Fig. 40, 
(upper diagram). It is not the arithmetical average tempera¬ 
ture difference, which would be the average vertical distance 
between the two dotted straight lines. For parallel flow, t m is 



Fig. 40. Two Cases of Heat Transfer. 


















Heat Transfer Appliances 


155 


obviously less than the arithmetical average. Its value for 
parallel flow or for cases where one fluid remains at constant 
temperature, or for cases in which there is no determinate flow, 
is given by 


id - t b - (t a ~ it') 


log* 


For counter flow, it is 


id - U 
t a - ib 


( 68 ) 


U - tb' - (ta - tb) 


log* 


ta' ~ td 

ta tb 


(68a) 


symbols being as in Art. 123. The derivation of these formulas 
furnishes an interesting exercise in calculus. 

In parallel flow, tb approaches t a as a maximum. In counter 
flow, it approaches the higher value t a Hence counter flow 
permits of higher ultimate temperatures of the heat-receiving 
fluid. On the other hand tb actually becomes equal to t a ' or t a 
only when the surface is infinite. A finite difference of tempera¬ 
ture always exists in practice at the outflowing point of the 
colder fluid (i.e., the right-hand end for parallel flow and the 
left-hand end for counter flow, in Fig. 40). With this difference 
fixed, t m is greater for parallel flow. The quantity of heat trans¬ 
mitted is proportional to t m . Hence when a given quantity of 
heat is to be transmitted, parallel flow is more economical of 
transmitting surface. 

Prob. 242. One fluid cools from 100° to 60° while the other 
is heated from 30° to 50°. What is the arithmetical average 
temperature difference? (Ans., 40°.) State the value of t m for 
parallel flow. (Ans., 30.9.) The direction of flow of the cooler 
fluid is now reversed and its final temperature changes to 68°. 
Find t m . (Ans., 30.9.) Compare the heat transfers in the two 
cases. (Ans., they are equal.) Compare temperature differ¬ 
ences at cold fluid outlet. (Ans., parallel flow, 10°, counter 
flow 32°.) 

Prob. 243. Should parallel flow or counter flow be used in 
the condenser of a C0 2 refrigerating machine? Why? 





156 Thermodynamics, Abridged 

125. Coefficient of Transmission. The general equation of 
transmission is 

Wa{Q,d — Qa) = Wb(Qb' - Qb) = pAtm, (69) 

where weights are in lb. per hr., A = surface in sq. ft., and 
p = coefficient of transmission. Obviously, p indicates 

THE QUANTITY OF HEAT (B.t.U.) TRANSMITTED PER HR. PER 
DEGREE OF MEAN TEMPERATURE DIFFERENCE BY EACH SQ. FT. 
OF surface. Few factors used in engineering are more variable 
than p. It depends on the two fluids, their condition and purity: 
and on the cleanliness and material of the transmitting surface. 
It is highest for transfer between liquids and wet vapors, as in 
condensers and feed water heaters: and lowest for transfer 
involving dry gases, as in superheaters and economizers. Besides 
the factors already mentioned, p varies notably with some power 
(0.3 to 0.6) of the velocity of fluid, or more strictly with the 
mass-flow (velocity X density). The values of p given in the 
accompanying table are merely rough averages. 

Prob. 244. A fan discharges 100 lb. of air per min. across 
coils through which water is being pumped. The water cools 
from 197.75° to 170.06°. The air is heated from 70° to 130°. 
What weight of water is pumped per min.? What amount of 
coil surface is necessary? Note: direction of flow is indeter¬ 
minate, hence use Equation (68). (Ans., 51 lb., 113 sq. ft.) 

Prob. 245. A feed water heater uses auxiliary exhaust steam, 
0.70 dry at 25 lb. pressure, amounting to 10 000 lb. per hr. 
Boiler feed water enters at 90°, its weight being 150 000 lb. per 
hr. The condensed steam is 10° warmer than the outlet feed 
water. What will be the outlet feed temperature? If counter 
flow is used, find t m and the necessary surface. (Ans., 140°, 
274 sq. ft.) 

Prob. 246. In Art. 47, 130 lb. of air are cooled 244° F., per 
minute, in the intercooler. Assume indeterminate flow, water 
entering at 70° and leaving at 110°. Find t m , and the necessary 
intercooler surface if p = 3. (Ans., t m = 199°, surface == 755 
sq. ft.) 


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Air blown at high velocity over coils. f Can be increased by improving brine circulation. 















































158 


Thermodynamics, Abridged 


126. The Steam Boiler. If the potential heat (heat of com¬ 
bustion) of the coal burned is B i, the heat evolved in the furnace 
is always some less quantity B 2 , the quotient B 2 /Bi being e F , 
the furnace efficiency. The heat evolved, B 2 , is available 
for transmission through the heating surface to the water and 
steam. Only a part, B 3 , is so transmitted, the remainder 
(B 2 — B 3 ) being lost up the stack. Then B 3 /B 2 = e s is the 
surface efficiency or transmissive efficiency and B 3 /B i = e^s 
is the efficiency of the whole boiler. 

For uniform coal, B\ varies in direct proportion to the amount 
of coal burned. Under good control, e F is about constant. 
Hence w T ith such good control B 2 is directly proportional to the 
coal burned. The heat B 2 exists in the form of a body of hot 
gas, and is proportional to the weight and temperature of gas. 
It is numerically equal to the product of weight of gas by its 
mean specific heat and by the difference between fire-room 
temperature and furnace temperature. The furnace tempera¬ 
ture and the mean temperature difference between gas and 
steam are scarcely affected by the weight of coal burned. The 
weight of gas is (or should be, in good operation) directly pro¬ 
portional to the weight of coal. 

The heat transmitted is B 3 = pAt m . In a given boiler, A is 
constant. Under good operation, as above described, t m is 
constant. Hence B 3 varies directly with p. Accept the mass- 
flow theory, i.e., p varies directly with the weight of gas, hence 
with the weight of fuel. Then B 3 varies directly with the weight 
of fuel. If e F varies, B 3 nevertheless varies directly with B 2 . 

This analysis loses sight of the fact that part of the transfer in 
a boiler is by radiant heat, with which the transmission varies 
not as the first power, but as the fourth power of the mean 
temperature difference. However, if t m near the furnace is 
constant, the radiant heat emission is constant. Hence the 
balance of heat transmitted is directly proportional to B 2 , or, 
at constant furnace efficiency, to J5i. 

This deduction is thoroughly confirmed by the straight line 
law illustrated in Fig. 41. The radiant heat emission is constant, 


Heat Transfer Appliances 


159 


and does not depend on the amount of coal burned. As a mathe¬ 
matical fiction, it may be regarded as represented by the inter¬ 
cept on the vertical axis ( oa , for the Denver). The balance 
of the heat is directly proportional to B i. Hence e s decreases 



Fig. 41. Navy Department Tests on Water-Tube Boilers—Quantities are 
per Square Foot of Heating Surface per Hour. 

with forcing, but more and more slowly, and the limit of boiler 
capacity is not easily reached. At very high rates of driving, 
e F decreases. 

If tests were carried down to zero weight of qoal burned, the 
lines could not, obviously, continue straight. This would imply 
negative radiant heat for the Nebraska. The imaginary 
straight portion (ab for the Denver) may, however, be utilized 
in calculation. See Probs. 247-250. 

A boiler “horse power” is equivalent to the transmission of 
34.5 X 970.4 = 33 479 B.t.u. per hr., or the evaporation of 34J 
lb. of water per hr. from and at 212° F. 












































160 


Thermodynamics, Abridged 


Prob. 247. For the Denver, in Fig. 41, hbw much heat was 
transmitted by pure radiation? (Ans., 860 B.t.u. per sq. ft. of 
heating surface per hr.) 

Prob. 248. When 20 000 B.t.u. coal value was supplied per 
sq. ft. of heating surface per hr., in the Denver, how much heat 
was transferred by radiation? (Ans., 860 B.t.u.) By ordinary 
transmission? (Ans., 11 800 B.t.u.) What was the boiler 
efficiency? (Ans., 0.633.) What was the hp. per sq. ft. of 
heating surface? (Ans., 0.377.) 

Prob. 249. State the boiler efficiency and hp. per sq. ft. of 
heating surface for the Denver when the coal equivalent is 
B i = 10 000 B.t.u. per hr. (Ans., 0.674 efficiency, 0.202 hp.) 



Fig. 42. Babcock and Wilcox Stationary Boiler with Superheater. 

Prob. 250. If t m = 1200, find p in Probs. 248, 249. Note: 
p applies to ordinary transmission only. (Ans., 9.83 and 4.90.) 

127. Superheaters. Superheaters (Fig. 42) are usually placed 
between the first and second “passes” of a water tube boiler* 
where the gas temperature is well above 1000°. The transfer 
equation is 











































Heat Transfer Appliances 


161 


W S k S (t 8 ' — t 8 ) = Wgkgitg ~ tg) = pAt m , (70) 

where w s = lb. of steam per hr. 

k s = specific heat of superheated steam between t s and t s ' 
at the given pressure. 

t s ' = temperature of steam after superheating, 
t s = temperature of saturated steam at the given pres¬ 
sure, 

w g = lb. of combustion gases passing the superheater per 
hr., 

kg = specific heat at constant pressure of combustion 
gases, say 0.24, 

tg' = temperature of gases approaching superheater, 
tg .=? temperature of gases leaving superheater, 
p = coefficient of transmission, around 3, 
t m = mean temperature difference, 

A = superheater surface, sq. ft. 

In very good operation, w g will vary directly with w s , the ratio 
Wg/ws being from 1.6 to 2.2. Hence 

ts' - ts 2 X 0.24 

tg' tg 0.48 ” 1 

when Wg/iCs = 2 and k s = 0.48 (rough average values). Thus 
the rise in temperature of the steam is about equal to the fall in 
temperature of the gas. As with boilers, p increases in direct 
proportion when w g increases: i.e., when the boiler is forced. 
Under these conditions, the equations can be satisfied only 
when the temperature ranges remain unchanged, because t m 
cannot increase when either temperature range increases. In 
practice, the ratios w g /w s and p/w g both vary somewhat and the 
amount of superheat varies in an irregular way with the rate of 
driving. 

Parallel flow should be used in superheaters to decrease 
the necessary surface. 

Prob. 251. A boiler of 1000 hp. uses feed water at 213° and 
generates dry steam at 200 lb. pressure. What weight of steam 
is produced per hr.? (Ans., 33 000 lb.) 




162 


Thermodynamics, Abridged 


Prob. 252. In Prob. 251, it is required to superheat this steam 
200°. Take k s = 0.548. The boiler evaporates 8.5 lb. of water 
per lb. of coal and the weight of gas is 17 lb. per lb. of coal. 
The gas after passing the superheater is at 1000° F. What is its 
temperature at entrance to the superheater? (Ans., 1230° F.) 
What is the mean temperature difference? (Ans., tfl0 o F.) The 
superheater surface? (Ans., 1970 sq. ft.) If the coal used 
contains 11 267 B.t.u. per lb., what is the combined efficiency 
of boiler and superheater? (Ans., 0.85.) Note: efficiencies 
above 0.80 are rarely attained in practice. 

Prob. 253. In Prob. 252, 10 per cent, of the heat is trans¬ 
mitted by pure radiation. Given t m = 1200, what should be 
the heating surface (not including superheating surface) of the 
boiler? (Ans., 2500 sq. ft.) 

128. Economizer. The definite equation is (see Art. 124), 

WwitJ — t W ) = 0.24 Wg{tg — tg) = 4 At m , (71) 

where w g /w w , as with the superheater, tends to remain around 2 in 
the best operation. Then the water is warmed about half as 
many degrees as the gas is cooled. Counter flow gives the 
highest final water temperatures, but economizer surface is 
costly and parallel flow may therefore be used. 

Prob. 254. Boiler feed water amounting to 10 000 lb. per hr. 
is to be heated in an economizer. Water enters at 90°, gas at 
500° and the ratio of weights of gas and water is 2.4. With 
parallel flow, the water leaves at 234°. Find the outlet gas 
temperature and the surface. (Ans., t g = 440° F., A = 1226 
sq. ft.) 

Prob. 255. If w w = 10 000, t w = 90, w g = 24 000, tj = 500, 
with counter flow, find the outlet gas temperature and the 
surface when tj = 270°. (Ans., t g = 425°, A = 1620 sq. ft.) 
Compare costs of installation and saving of heat in Probs. 254 
and 255. (Ans., the counter flow economizer costs 32 per cent, 
more but it puts 25 per cent, more heat into the feed water.) 


Heat Transfer Appliances 


163 


129. Surface Condenser, Feed Water Heater. In surface con¬ 
densers (Fig. 43) and feed water heaters (Fig. 44), some of the 
factors affecting the value of p have been established. For usual 
tubes when clean, p = 400 Vw, where u = water velocity, ft. per 
sec. Very badly fouled tubes will give half this value of p. 


.ExhaustSteam (° 0uikt 

WaterOutlet' 


Tube 

Sheet) 



& 

V Water 
Inlet B 

Section A-A 


From 

Inlet 


.. I Path of Circulating} 
iCylinder { Water 
Drain ir Pump Su ction 

Section B-B 


Fig. 43. Surface Condenser Showing Path of Water and Steam and General 
Arrangement of Baffle Plates and Tubes. 


Vulcanizing (coating) admiralty tubes decreases p nearly 85 
per cent. If there is much air in the steam to be condensed, 
p will be reduced. Air may enter through leaks in vacuum 
piping. Fresh feed water also carries air. High water velocities 
increase p and therefore decrease the necessary size of heater or 
condenser. They also increase the power consumption of the 
pump which handles the water. This last point is not of much 
importance in condensers, where circulating water pressures are 
low: but the water passing through feed water heaters is under 
full boiler pressure and excess pressure should be avoided. With 
high water velocities, excess pressures are necessary to overcome 
friction. For feed water heaters, use u — 6 or 7. 

For condenser or heater, with clean tubes of usual materials, 

w s (hi + xiLi — h 2 ) = w w (U — h) = 400 ^uAt m , (72) 























164 


Thermodynamics, Abridged 





Fig. 44. Marine Feed Water Heater. 











































































































Heat Transfer Appliances 


165 


where i and 2 are initial and final steam conditions, 3 and 4 initial 
and final water conditions. Flow is usually indeterminate, hence 
Equation ( 68 ) is used for t m : this equation becoming, in present 
symbols, 


tm — 


tl ~ ts — (£ 2 — £ 4 ) 



Prob. 256. A condenser handles 17 400 lb. of steam per hr. 
at 0.696 lb. pressure, 0.85 dry. The condensed steam is dis¬ 
charged at 80°. The circulating water enters at 60° and leaves 
at 74°. Find the weight of water. (Ans., 1 113 000 lb. per hr.) 

Prob. 257. The condenser in Prob. 256 has 600 tubes in 
each “pass”—see Fig. 42. These each have an internal cross- 
sectional area of 0.219 sq. in. If the weight of water is 64 lb. 
per cu. ft., find the water velocity through the tubes. (Ans., 
4.82 ft. per sec.) 

Prob. 258. In the foregoing, find the values of p, t m and A. 
(Ans', p = 878, t m = 14.9, A = 1193.) 

Prob. 259. In Probs. 256-258, each lineal ft. of tube gives 
0.1635 sq. ft. of surface. There are two passes. What is the 
length of tubes? (Ans., 6.5 ft.) 

In feed water heaters, p is sometimes increased by the use 
of retarders or by employing corrugated tubes. Sometimes a 
double corrugated tube is used, the water passing through the 
narrow space between the inner and outer walls. The object 
of increasing p is to decrease surface and thereby to save space 
and cost. Many special forms of heater increase p without 
decreasing total bulk or cost, because the surface is of such 
shape as to be of large bulk per unit of transmitting area. Con¬ 
densers generally have plain straight tubes. 

130. Evaporator. Essentially, an evaporator is a shell con¬ 
taining salt water in which there is immersed a coil containing 
steam. Pure steam is driven off from the sea water and later 
condensed, thus furnishing a supply for boiler feed and domestic 




166 


Thermodynamics, Abridged 


purposes. Fig. 45 shows considerable elaboration on this. The 
salt feed is first used as cooling water in a small condenser or 
“distiller.” The bulk of it is then wasted, but a part feeds the 
evaporator shell. Its temperature having been somewhat raised 


To main engine 
condenser or hot welf 


Separator\ 
Fromboilerthroughpressure X 3 L 3 h 3 {\\ 
reducing valve or from receiver 
of compound engine —=>. 


Wr 


SaltWater 
/Supply 
{ (w 2 +w 4 +w 5 +w 6 ) 

^ k- Distiller 
-Coils 




Evaporator— ->g 




ns 


Coils 



y h s 

J 3 Vrain (Brine) 
"Baffle 




Dripf W s "BlowDown 

Ik 


-Drinking 

Water 


h 4 \L(K+K+K) 


Waste 


Fig. 45. Evaporator. 


at the distiller, less heat will now be required to evaporate than 
would otherwise be needed. In the evaporator the salt water is 
boiled: and, if some of the coil is unsubmerged, slightly super¬ 
heated. As the sea water is vaporized, the salt left behind con¬ 
tinually strengthens the brine in the evaporator shell. Hence the 
contents must be frequently or steadily blown down to keep the 
salinity within reasonable limits. 

The vapor leaving the evaporator passes through a separator 
where any moisture is baffled off. It may then go to the distiller 

































Heat Transfer Appliances 


167 


and be condensed for use as drinking water, etc. If it is desired 
to use the condensed pure water for boiler make-up, it can be 
carried directly to the main engine condenser. The whole 
evaporator will then carry a vacuum: instead of “blowing” 
down, a pump must be used to withdraw concentrated brine: 
and the heat in the vapor (though not the vapor itself) will be 
discharged outboard (and therefore wasted) with the condenser 
circulating water. A better plan is to carry the vapor to the main 
hot well for condensation. Its heat as well as its substance is 
then transferred to the boiler feed. 

Assume the distiller to be in use. Then with symbols as in 
Fig. 45, 

Qi = heat supplied by live steam at evaporator 

= Ws(hs + x&L$ — h\)> 

Q 2 = heat absorbed at evaporator 

= (^5 T" W 4)(^3 + x 3 L 3 — h 2 ) w^(h^ — 7 * 2 )? 

Qi = Q 2 + radiation at evaporator, 

Qs = heat discharged at separator = wji 3 , 

1 — Xz 

w 4 = w 5 -, 

£3 

Q 4 = heat rejected at distiller = w b (H 3 — h b ), 

Q b = heat absorbed at distiller = (w 2 + + w b + Ws)(h 2 - ~ hi), 

Q a = Q b + radiation at distiller. 

A test on a small machine gave the following values: p 8 = 80, 
x 8 = 0.99, U = 292.7, w 8 = 2800, p 3 = 25, z 3 = 0.99, t 2 = 126.15, 
w b = 2340, wq = 200, U = 237.8, h = 209.55, h = 80. The coil 
transmitting surface in the evaporator was 70 sq. ft. Weights 
are in lb. per hr. The following are tabular properties: 
h = 282, L 8 = 900.3, h = 262.1, h = 208.4, L 3 = 952, 
h 2 = 94, h = 206.1, h = 177.5, h = 48.03. 

Then 

W\ = 2340 X -gV — 23.6, (^5 ~f ~ ^ 4 ) = 2363.6, 

Q x = 2800 (282 + 09 9 X 900.3 - 262.1) = 2 552 352 B.t.u., 

Q 2 = 2363.6 (208.4 + 0.99 X 952 - 94) + 200(206.1 - 94) 

= 2 522 420 B.t.u. 


12 





168 Thermodynamics, Abridged 

Then the radiation loss at the evaporator is Q\ — Q 2 = 29 932 
B.t.u., 

Q 4 = 2340 (208.4 + 952 - 177.5) = 2 300 000 B.t.u., 

Q b = ( W2 + 2363.6 + 200) (94 - 48.03) 

= 45 . 97^2 + 118 000 B.t.u. 

Disregarding radiation at the distiller (it should be very small), 
put §4 = Q 5 . Then w 2 = 47 400 and the weight of circulating 
water pumped to the distiller is 47 400 + 2563.6 = 49963.6 lb., 
the greater part of which is wasted. 

The heat transmitted at the evaporator was 2 522 420 B.t.u. 
per hr. The mean temperature difference (counter flow) was 
t m = 114. The value of p realized was then 2 522 420 -s- (70 
X 114) = 316. Assuming the same value to have been realized 
at the distiller, with indeterminate flow and t m = 145, the neces¬ 
sary surface there is 2 300 000 -r- (145 X 316) = 50 sq. ft. 

Higher values of p may be realized for a short period, but the 
accumulation of scale is rapid. 

131. Multiple Effect Evaporator. In Art. 130, 2340 lb. of dis¬ 
tilled water were produced per 2800 lb. of live steam condensed, 
or 0.833 lb. of water per lb. of steam. A large waste of heat 
occurred at the point w 2 , h 2 , Fig. 45. If the vapor leaving the 
separator had been conducted to the coil of a second evaporator, 
in which the shell pressure was well below 25 lb., additional 
evaporation might have been produced there. Such an arrange¬ 
ment would constitute a multiple effect machine. 

Thus, in Art. 130, the vapor leaving the separator is at 25 lb. 
pressure and dry. Conduct to a second evaporator in which 
the shell pressure is 10 lb., the supply temperature 90° and the 
emerging vapor dry. Assume the coil drip to be at 230.6°. 
The weight of vapor delivered to this coil is w 5 = 2340 lb. The 
heat it delivers is 2340 {H 2 5 — h 2 30 . 6 ) = 2340 (1160.4 — 198.8) 
= 2 250 000 B.t.u. The heat required to vaporize 1 lb. of 
liquid fed to the shell is Hi 0 — h 90 = 1143.1 — 58 = 1085.1 
B.t.u. The weight vaporized is then (ignoring radiation and 
blow down losses) 2 250 000 -f- - 1085.1 = 2070 lb. In the two 


Heat Transfer Appliances 


169 


effects combined, the distilled water produced per lb. of original 
steam condensed is (2070 + 2340) -f- 2800 = 1.57. By using 
three effects, the economy might have been still further im¬ 
proved. Actual rates range up to 3.5 or 4 lb. of vapor per lb. 
of steam with six effects. The number of stages is limited by 
the total range of steam pressures. There must be a consider¬ 
able temperature difference (and hence pressure difference) 
between coil and shell. The use of exhaust steam for the first or 
“primary” coil would make the complication and increased cost 
of multiple effect apparatus unnecessary: but with exhaust 
steam the mean temperature difference is low, the surface large 
and the apparatus expensive. Recent capital ships provide for 
the use of exhaust, but the amount of exhaust available is in¬ 
adequate for continuous running. Weight limitations also pre¬ 
clude the exclusive use of exhaust steam. 

When several stages are used, the latter ones are employed not 
for actual vaporization, but for heating the incoming salt water, 
just as the distiller heats it in Fig. 45. Matters are so adjusted 
that there is no surplus of water to waste between effects. As 
with heat transfer in general, parallel flow economizes surface 
while counter flow leads to a higher final temperature at the 
distiller discharge and hence permits of more stages (and more 
economy) with a given pressure range between primary steam 
supply and condenser. In parallel flow, the coldest salt water 
enters the first (highest pressure) stage: in counter flow, it enters 
the last (lowest pressure) stage. 

Prob. 260. In Art. 130, check the values of t m at evaporator 
and distiller. 

Prob. 261. In Fig. 45, w 8 = 2000, p 8 = 75, x 8 = 1.0, U = 
302°, radiation at evaporator = 20 000 B.t.u. per hr., t 2 = 153.01, 
p 3 = 30, x 8 = 1.0, w 8 = 0. Find weight of distilled water pro¬ 
duced per lb. of steam supplied. (Ans., 0.86 lb.) 

Prob. 262. In Prob. 261, h = 90, U = 205.87. Radiation at 
distiller = 14 000 B.t.u. per hr. Find weight of salt water 
pumped through the distiller, and weight of this water that is 
wasted. (Ans., 26 900 lb. and 25 180 lb.) 


170 


Thermodynamics, Abridged 


Prob. 263. With p = 300, find surface necessary at evapor¬ 
ator and at distiller in Probs. 261 and 262. (Ans., 62 sq. ft. 
and 53 sq. ft.) 

Prob. 264. In the foregoing, instead of using a distiller, the 
vapor from the evaporator is carried to a coil in a second evapor¬ 
ator. The pressure in the shell is here 20 lb. The coil drip is 
at 240.1°, the salt water supply at 141.52° and the vapor pro¬ 
duced is dry. If radiation here is 18 000 B.t.u. per hr., and 
p = 300, find the necessary surface in this second effect. (Ans., 
104 sq. ft.) How much distilled w^ater is produced per lb. of 
original steam by the whole machine? (Ans., 1.635 lb.) 

132. Radiation Loss from Steam Pipes and Vessels. For 

bare pipes or steam vessels with little or no air circulation and 
with only such steam circulation as is due to condensation, 
p = 1.8, as for steam radiators in the table of Art. 125. The 
value may rise to 5 or more if the steam temperature is very high 
and if there is a flow of steam. Commercial coverings lead to 
values of p below 0.5. 

Prob. 265. A 10-in. pipe (lOf in. outside diameter, 81.55 
sq. in. internal cross-sectional area) carries steam initially dry 
at 200 lb. pressure, a distance of 300 ft. The steam velocity 
entering the pipe is 8000 ft. per min. The surrounding air is at 
50° F. Take p = 3, the pipe being bare. How much heat is 
lost per hour? What weight of steam flows per hr.? What 
proportion of its original heat (above 32°) does the steam lose? 
Disregarding drop of pressure, what is the dryness of the steam 
at the far end of the pipe? - (Ans., 841 000 B.t.u. lost per hr.: 
118 400 lb. flow per hr.: loss = 0.59 per cent.: dryness = 0.99.) 
Note: the loss of heat goes on whether the pipe is delivering 
steam at full capacity or not: hence the percentage of loss is 
great when the delivery of steam is small. 


CHAPTER VIII 


FLUID FLOW AND THE STEAM TURBINE 
Flow of Steam 

133. New Assumption. Thus far, Equation (3) has been our 
basis. There is one effect of heat not recognized in that Equa¬ 
tion as heretofore applied, i.e., the production of kinetic energy. 
Where the addition of heat produces appreciable fluid velocities, 
the general equation must be written 

H=T+/+W+7, (73) 

where V stands for that part of the heat H which is converted 
into kinetic energy or velocity energy. 

If 1 lb. of steam works from condition i to condition 2 without 
reception or emission of heat, the total energy bring constant, 
and U representing velocity, 

Ui U 2 2 

(T+ /+ W )!+ 2^= 7 + + 

Now (T + /+ W) may be taken as the total heat of steam 
above 32°, whether the steam be wet, dry or superheated. 

Designating such total heat by Q, 

77SQ 1 + U ^= 778Q,+ 

In most cases, U 2 is negligibly small* as compared with U 2 . 
Hence, nearly, 

U 2 2 = 778 X 2 g(Q 1 - Q 2 ), 

17, = 224 VQi- Q 2 , (74) 

where TJi is the final velocity in ft. per sec. Values of Qi and Q? 
are to 'be taken as for adiabatic constant-entropy expansion 

* In a steam turbine, lh is the velocity of steam at the entrance to the 
nozzle and Ui is the velocity at the nozzle outlet. The ratio of Ui to U\ 
may be about 20: hence that of Ui 2 to Ui 2 is about 400. 

171 



172 


Thermodynamics, Abridged 


from the total heat-entropy diagram, Fig. 26. The corresponding 
value of U 2 is the velocity attained by expansion in an ideal 
frictionless nozzle. 

Prob. 266. Dry steam at 200 lb. pressure expands adiabat- 
ically to 1 lb. pressure. What is the ideal velocity attained? 
(Ans., 4100 ft. per sec.) 

134. Effect of Friction. In all actual nozzles, there is a fric¬ 
tion effect due to the rubbing of steam against metal. If there 
is sufficient friction, as when the orifice is very small, the velocity 
may be entirely destroyed. What then becomes of the heat 
developed by friction? It can only go back into the steam. 
The law of such process (when the velocity is wholly destroyed) 
is, Qi = Q 2 : the total heat is constant. The expansion valve 
operation in vapor refrigeration, path def, Fig. 36, illustrates 
this. No heat has been absorbed or emitted from without: 
but the entropy has changed. Heat has been converted into 
velocity, and then reconverted, by means of friction, into heat. 

In considering the injector (Prob. 240), nothing was said of 
velocity. The heat lost by the steam is first converted into 
velocity. There then follows an exchange of energies between 
water and steam, the kinetic energy is greatly reduced and the 
reduction is applied to increase the heat existing as such. In 
Prob. 240, it was assumed that the total heat was the same after 
mixture as before. This is not strictly true. A small part has 
become kinetic energy, represented by the moving body of warm 
water on its way to the boiler. At a boiler pressure of even 200 
lb., each lb. of water must have velocity (kinetic energy) merely 
sufficient to overcome a head of about 460 ft., or the necessary 
kinetic energy is 460 ft. lb.—little more than half of 1 B.t.u. 
The error in Prob. 240 is therefore exceedingly small. 

135. Throttling Calorimeter. This device consists essentially 
of a very small orifice through which steam is expanded. The 
velocity is wholly reconverted to heat, so that the law of the 
process is Qi = Q 2 . Inspection of Fig. 26 will show that if the 
total heat remains constant while the pressure decreases, the 


Fluid Flow and the Steam Turbine 


173 


steam becomes drier and is finally superheated. The process 
is represented on Fig. 26 by a horizontal line drawn from left to 
right. The calorimeter is used to determine the quantity of 
moisture in steam. The pressure is measured on the inlet side, 
and pressure and temperature are noted on the outlet side. 
Then if the amount of expansion is sufficient to produce super¬ 
heat at the outlet, 

(h + xL)^ {H+ k(t'-t)} 2 . (75) 

Values of hi, L\, t 2 and H 2 are taken from the table, for the pres¬ 
sures noted. The observed value of t 2 then indicates the 
original dryness of the steam. The method fails when V ^ t, 
i.e., when no superheating occurs. This may result either from 
insufficient expansion or from the use of steam too wet to begin 
with. If a measured quantity of heat Q is added to each lb. of 
steam on the high-pressure side, the range of the instrument can 
be increased: 

Q+ (h+ xL)i = {H + W - t) } 2 . (76) 

Problems may readily be solved by the use of Fig. 26, noting 
that a constant heat process implies a horizontal path. 

Prob. 267. Steam at 100 lb. pressure is expanded to 15 lb. 
pressure in a throttling calorimeter. The temperature is then 
found to be 216°. What was the original dryness? (Ans. 
0.961.) 

Prob. 268. What is the wettest steam at 100 lb. pressure 
that can be tested in an ordinary throttling calorimeter using 
pressures as in Prob. 267? (Ans., 0.959 dry.) 

Prob. 269. Steam at 100 lb. pressure has added to it 100 
B.t.u. per lb. When expanded to 15 lb. its temperature is 
found to be 263°. What was its dryness? (Ans., 0.874.) 

136. Velocity and Flow in Actual Nozzles. In Fig. 46, let a 
be the initial condition of the steam, determined by its pressure 
and quality. Let h indicate its condition at the same entropy 
but at some lower pressure, reached after expansion in a nozzle. 
Suppose friction to be such that 10 per cent, of the velocity 


174 


Thermodynamics, Abridged 


energy is re-converted to heat energy. Then lay off hk = 0.10 
X ah = 0.10(Q a — Qh )• The heat content after expansion is 
Qk, but the pressure is ph'- hence the condition of the steam is 
indicated by the point j, found by drawing kj horizontally. 



Thus, if p a — 200, superheat at a = 200°, and p h = 180: Fig. 
26 gives Q a = 1308, Q h = 1297. Then hk = 0.10 X 11 = 1.1, 
Qk = Qj = 1297 + 1.1 = 1298.1. Fig. 26 now shows the steam 
to be 186° superheated at the point j. The velocity due to 
expansion from 200 lb. to 180 lb. is 224Vl308 — 1298.1 = 705 
ft. per sec. Note that the heat drop is Q a — Qj instead of 
Qa — Qh, in an actual nozzle with friction. 

Fig. 46a plots resulting velocities (J7 2 ) for these conditions, 
with various pressures attained during expansion. The specific 







Fluid Flow and the Steam Turbine 


175 


volumes, v, of steam are also plotted on this diagram. Thus 
for an expanded pressure of 180 lb., with 186° of superheat, the 
specific volume is (Art. 83) 3.3 cu. ft. 



Fig. 46a. Flow in Nozzle. 


If Wo = weight of steam discharged, lb. per sec., the cross- 
sectional area of the nozzle, at a point where the pressure is p, 
is vwo -f- XJ 2 sq. ft. Values of v/U 2 are plotted in Fig. 46a: 
which is based on an initial pressure of 200 lb., 200° initial super¬ 
heat and 10 per cent, reconversion of kinetic energy to heat. 

Suppose the “ nozzle ” is merely a hole in a plate. In this 
as in any other case, the area of cross-section is mathematically 
infinite on the high-pressure side, for here C7 2 = 0. In practice, 
this side is simply well chamfered. If the pressure of discharge 
is 180 lb., the area for 1 lb. discharge per sec. is (from Fig. 46a) 
0.0047 sq. ft. If the discharge pressure is 140 lb., the area per 
lb. discharged per sec. is 0.0031 sq. ft. Suppose the actual area 
to be 1 sq. ft. Then the weights discharged per sec. are 1 -f- 
0.0047 = 212 lb. if the discharge pressure is 180 lb., and 1 -F 
0.0031 = 323 lb. if it is 140 lb. If the discharge pressure is 116 
lb., the weight is a maximum, and is 1 -F 0.00291 = 344 lb. per 
sec. The diagram shows clearly that the area necessary to dis¬ 
charge a given weight is a minimum when the discharge pressure 




































































176 


Thermodynamics, Abridged 


is 116 lb. Conversely, the weight discharged through an orifice 
of given area is a maximum when the discharge pressure is 116 lb. 
A further decrease in discharge pressure does not increase the 
weight flowing. The discharge pressure at which maximum 
weight of flow is realized is called the critical pressure. For 
steam, the critical pressure is always 58 per cent, of the initial 
pressure (it is 116 lb., in this case). For air, it is 53 per cent. 

Now consider an actual nozzle designed for 200 lb. initial 
pressure and 200° initial superheat, with 10 per cent, recon¬ 
version. It will have a well rounded entrance. If it is a con¬ 
verging nozzle, it will decrease in cross-sectional area from 
inlet to outlet. Suppose, in such case, the outlet area to be 1 
sq. ft. The weight of flow will depend on the outlet pressure and 
will be (as above) 212 lb. if the discharge pressure is 180 lb., 
323 lb. if the pressure is 140 lb., and 344 lb. if it is the critical 
pressure of 116 lb. The corresponding outlet velocities, also 
from Fig. 46a, are 705, 1290 and 1590 ft. per sec. If the dis¬ 
charge pressure is the critical pressure (116 lb.) then the pressure 
will be 140 lb. at a point in the nozzle where the cross-sectional 
area is 0.0031/0.00291 — 1.065 sq. ft., and the velocity at that 
point will be 1290 ft. per sec; 

If the discharge pressure is less than the critical pressure, the 
nozzle should first contract to a throat and then diverge 
(diverging nozzle). Thus in Fig. 46a, the pressure in the throat 
(narrowest section) is 116 lb. The ordinate of the nozzle area 
curve is there 0.00291. At a point beyond the throat, where the 
cross-sectional area is 1.047 sq. ft., the ordinate corresponding 
is 1.047 X 0.00291 = 0.00305, so that the pressure at that 
point is 80 lb. and the velocity is 2030 ft. per sec. Note that 

THE DIVERGENT PORTION OF THE NOZZLE DOES NOT AFFECT THE 
WEIGHT OF FLOW, BUT ONLY THE FINAL (OUTLET) PRESSURE 

and velocity. The weight of flow is determined by the throat 
area, and the throat pressure is the critical pressure; if the final 
pressure is less than the critical pressure. 

The reason for these relations is suggested by Fig. 46a. As 
expansion proceeds, pressure falls and volume and velocity both 


Fluid Flow and the Steam Turbine 


177 


increase. The volume at first increases more slowly than the 
velocity, and afterward more rapidly. The stream of steam 
reaches its minimum cross-section where the quotient of volume 
by velocity is a minimum. 

The long equation of Art. 83 need be used for specific volumes 
only when the steam is superheated. After the expansion 
passes the saturation curve (Fig. 46), the dryness may be read 
from Fig. 26 and the corresponding wet steam volume computed 
directly. 

Approximate formulae for flow in lb. per hour are 
w = 60 ^ forsaturatedsteam 


w = 


1+ o!000650a - t) for superheated steam ! 


(77) 


where A t = throat area in sq. in., (t a — t) = superheat. These 
apply only when the discharge pressure is less than 0.58 the 
initial, and are based on a reconversion of about 7 per cent, 
instead of 10 per cent. For air, when the discharge pressure 
is equal to or less than 0.53 the initial, 


w = 


1900 y a At 

W a 


(78) 


The velocity in the throat is then that of sound. 


Prob. 270. (a) Steam at 200 lb. pressure, 200° superheated, 

is discharged through a converging nozzle with 10 per cent, 
reconversion to a chamber where the pressure is 40 lb. The 
outlet area is 0.1 sq. ft. What is the pressure in the outlet? 
(Ans., 116 lb.) What weight of steam will be discharged? 
(Ans., 34.4 lb. per sec.) What is the outlet velocity? (Ans., 
1590 ft. per sec.) (6) The discharge pressure is changed to 140 
lb. State outlet pressure, velocity and weight discharged. 
(Ans., 140 lb., 1290 ft. per sec., 32.3 lb.) (c) A diverging 
nozzle having the same reconversion is used for the same initial 
steam conditions. It discharges 2000 lb. per hr. with a discharge 
pressure of 1 lb. Find pressure, velocity, specific volume and 





178 


Thermodynamics, Abridged 


area at the throat and at the outlet. (Ans., at throat, 116 lb., 
1590 ft. per sec., 4.6 cu. ft., 0.00161 sq. ft. At outlet, 1. lb., 
4130 ft. per sec., 290 cu. ft., 0.0389 sq. ft.) ( d ) Under case (c), 
state values of total volume per hr. -t- velocity in ft. per 
sec. at throat (Ans., 5.79), outlet (Ans., 140), and at a point in 
the expansion where the heat content is 1274.7 and the specific 
volume is 4 (Ans., 6.2). 

Prob. 271. Check Prob. 270, Case (c), as to weight flowing, 
by Equation (77). (Ans., Equation gives 2100 lb. per hr.) 
What weight of dry steam at 200 lb. pressure would flow through 
the same nozzle? (Ans., 2400 lb.) What weight of air at this 
pressure and at 40° F.? (Ans., 3900 lb.) What would be the 
throat pressure with air? (Ans., 106 lb.) What is the throat 
velocity with air? (Ans., 1180 ft. per sec.) 

Prob. 272. Determine throat and outlet areas of a nozzle to 
discharge 500 lb. of dry steam at 200 lb. pressure, per hr., if 
the discharge pressure is 2 lb. and the reconversion is 10 per cent. 
(Ans., 0.0528 and 0.768 sq. in.) 

137. Nozzle Efficiency. The percentage of reconversion is 
determined by the design of the nozzle. If there were no fric¬ 
tion, there would be no reconversion and the nozzle efficiency 
would be 100 per cent. Good efficiency requires smooth interior 
finish, a well rounded entrance and a proper angle (15° or so) 
between the diverging sides. Most important, the outlet area 
and throat area .should be in suitable ratio for the pressures 
used. If either pressure (inlet or outlet) differs from that con¬ 
templated in design, eddies will be produced and a loss of effi¬ 
ciency will result. If e = nozzle efficiency, 1 — e = proportion 
of reconversion and U 2 = 224 ^e(Q a — Qb ), Fig. 46. The value 
of U 2 varies, thus, not with e but with Ve. Values of e up to 
0.94 may be expected. Note that e = U 2 -f- 50 056 (Q a 
- &). 

Prob. 273. The nozzle efficiency is 0.94. How much is 
velocity reduced by friction? (Ans., 3 per cent.) 



Fluid Flow and the Steam Turbine 179 

Simple Impulse Turbines 

138. Simple Velocity Diagram. The small sketch in Fig. 466 
indicates the operation of a single-stage impulse turbine, like 
the De Layal. The wheel W carries curved blades or buckets, 
B and revolves with its shaft S. The nozzle N directs the jet 
at an acute angle (about 20°) with a plane of rotation. 



The larger diagram shows the velocities involved. This 
diagram lies in an imaginary plane which, in the upper of the 
sketch views, passes through the axis of the nozzle and is per¬ 
pendicular to the plane of the paper. 

Draw gb representing the line of motion of the bucket. Draw 
ab representing the direction of motion of the steam jet, the two 
lines intersecting at the nozzle angle a. Let the lengths ab 
and be represent the magnitudes of steam and (peripheral) 
wheel velocities. Now the relative* steam velocity (velocity 

* Absolute and Relative Velocities. (1) A ship sails West at 10 knots 
speed. The wind blows west at 14.17 knots (absolute velocity of wind). 
The relative velocity of the wind with respect to the ship is 4.17 knots. 












180 


Thermodynamics, Abridged 


as it would seem to be to an observer moving with the wheel) 
is the resultant ac. This is deflected by the bucket to the direc¬ 
tion cd. Friction and eddy losses make the ultimate relative 



Fig. 47. Impulse Turbine Triple Expansion. 

exit velocity less than the initial relative velocity, the relation 
of the two being as in Fig. 48. Let ch denote the magnitude of 
the relative exit velocity. Combine this with the wheel velocity 
(hj == be) to obtain the absolute or true velocity, cj, of steam 
leaving the buckets. 

Force is change of momentum. For 1 lb. of steam, the initial 
momentum is ab/g. The momentum in the direction of 
movement is bg/g. The final momentum in the same direction 
is — ck/g. The negative sign indicates that this is a rearward 
momentum. The corresponding force is a reactive impulse, or 

westward. (2) The wind blows east at 14.17 knots. The absolute wind 
velocity is 14.17 knots, the relative velocity is 24.17 knots, east. (3) The 
wind blows south at 10 knots absolute velocity. The relative velocity 
will be 14.17 knots, southeast. (4) The wind blows southwest at 14.17 
knots absolute velocity. The relative velocity is 10 knots, south. In each 
case, the relative velocity is directed aft as compared with the absolute velocity. 











Fluid Flow and the Steam Turbine 


181 


reaction, while the force corresponding with the momentum 
bg/g is an impulse. The change of momentum, or force im¬ 
pelling the bucket, is ( bg + ck)/g. The speed of the bucket is be • 



Fig. 48. Bucket Friction. 


Work is force times speed. Hence the work obtained per lb. 
of steam is 

he 

W=j (bg+ ck) ft. lb. 

_ bcibg + ck) _ bc(bg + ck) 

32.17 X 778 25028 \ ' 

Prob. 274. Steam at 150 lb. pressure, and dry, expands to 
1 lb. pressure with a nozzle efficiency of 0.93. Find the resulting 
velocity. (Ans., 3850 ft. per sec.) 

Prob. 275. With steam velocity = 3850, bucket velocity 
= 600, a = 20°, construct to some convenient scale a velocity 













































































182 


Thermodynamics, Abridged 


diagram like that of Fig. 466, the relative angles acg and deg being 
equal. Measure the necessary lines and state the work done per 
lb. of steam. (Ans., 125 B.t.u.) 


139. Kinetic Efficiency. The efficiency of conversion of kinetic 
energy into work at the buckets is what we may call the 
KINETIC EFFICIENCY, 


&Jc 


w + 


uj 


2g X 778 


50 056JF 
U ab 2 


(79) 


It is easily shown that the maximum value with equal relative 
angles is attained when be = %bg, and that that value is 




where F = bucket friction coefficient = ch/ca = ch/cd, to be 
taken from Fig. 48. If there is no bucket friction, F = 1 and 
e km&x = (bg/ab) 2 . If a = 0, then the point k would coincide 
with the point / and e k max would be 100 per cent, when be — \ba. 
Note that a is a necessary evil and that the ideal performance is 
realized when cj is vertical. 

Maximum kinetic efficiency occurs when the peripheral 

SPEED OF THE BUCKETS IS HALF THE ROTATIVE COMPONENT OF 

the steam speed. High peripheral speeds are essential to 
efficiency. 

Prob. 276. Compute ek in Prob. 275. (Ans., 0.42.) 

Prob. 277. With a = 0°, Fig. 47, ged = 0°, F — 1.0, what 
would have been the best value of be, W and ek for Prob. 275? 
(Ans., be = 1925 ft. per sec., W = 297 B.t.u., ek — 1.0.) 

Prob. 278. In Prob. 275, for a = 20°, F — 1.0, state the 
best value of be and the resulting values of W and ek, the relative 
angles being equal. (Ans., be = 1810 ft. per sec., W = 261 
B.t.u., ek — 0.88.) 

Prob. 279. With the bucket friction and value of a assumed 
in Prob. 275, and with equal relative angles, state the best values 
of be, W and e k - (Ans., be = 1810, e k = 0.76, W = 226 B.t.u.) 




Fluid Flow and the Steam Turbine 


183 


Fig. 49 shows the variation in W with bucket speed, be, for 
the conditions of Prob. 275, except that the friction factor F is 
taken constantly at 0.75. The variation of F is such that the 
work is actually a little greater than is indicated by the curve, 
at high speeds, and a little lower at low speeds. Note that the 
curve of Fig. 49 represents e k (to another scale). 


.«£ 



Peripheral Speed of Buckets,ft per sec. 


Fig. 49. Effect of Bucket Speed on Work at Buckets and on Kinetic 
Efficiency. 


140. Primary Difficulties in Turbine Design. Let u (= be, 

Fig. 466) denote peripheral speed in ft. per sec., N = r.p.m., 
d = wheel diameter (pitch diameter) in ft. Then irdN — 60 u or 
dN = 19.Dl For high values of u, either d or N must be high. 
The smaller the capacity, the smaller preferably, is d: hence N 
is high in small machines. Flexible shafts and reduction gears 
may be essential in turbines of this type. The largest values 
of u in actual use are around 1300; of N, around 24 000. 

The capacity of an impulse turbine is determined by the nozzle 


13 






































184 


Thermodynamics, Abridged 


area and steam conditions rather than by any dimension of the 
wheel. The height of the buckets should somewhat exceed the 
nozzle outlet diameter, to prevent lateral spreading of steam. 
The wheel runs in a chamber containing steam at condenser 
pressure, and (in this type) need not make a close fit with its 
casing. No high pressure steam is carried beyond the nozzle. 
The pressure is constant across the face of the buckets. 

Prob. 280. In Prob. 275, what is the pitch diameter of the 
bucket wheel if it makes 10 000 r.p.m.? (Ans., 1.146 ft.) 

141. Turbine Efficiency. The turbine must be regarded as 
operating in the complete expansion (Rankine) cycle: Art. 80. 
Its ideal efficiency is therefore to be computed as in Equations 
(45), (49). We have seen that the conversion of heat into 
kinetic energy is accompanied with a nozzle friction loss, and 
that the conversion from kinetic energy into work at the buckets 
is accompanied with a further loss. Let e T = efficiency of ideal 
cycle, e N = nozzle efficiency, e & = kinetic efficiency. Then the 
efficiency from steam to work at the buckets is X e N X e*. 
Further loss now occurs. 

The revolving disk and the rows of buckets both undergo 
friction against the steam. For wheels under 36 in. diameter 
revolving in a chamber containing dry steam, empirically, 

L d = d 2 u 2 9 4400v B.t.u. per hr., I 

Lb = dh l - 2b u 2 -* -5- 990a B.t.u. per hr., J 

where a = specific volume of dry steam at nozzle outlet pressure, 
h = radial height of buckets, in., L d = friction loss at disk, 
Lb = friction loss for one row of buckets. For m rows of buckets 
on one disk, the total rotation loss per disk is L d + mLb . 
This is to be multiplied by the factors of Fig. 50 if the steam is 
(as usually) not dry in the wheel chamber. In Fig. 46, the steam 
enters the chamber as it leaves the nozzle, at the state d. It 
leaves the chamber at a state h, defined by Q a — Qh = W, the 
pressure at h being the same as at d. The mean of the drynesses 
at d and h determines the abscissa of Fig. 50. 


Fluid Flow and the Steam Turbine 


185 


o 

oo 

to 

§1 

~cu 

■=S 

.cr, 


to 
to . 


c: 

o 


o 

a: 


o 

”o 


CD 

> 


(D 














i 













f 












1 

r 












1 













1 


— ^ 











r 













1 












1 

f 












7 













i 



) 









^7 

/ 




> 









/ 

/ 












/ 

/ 












J 

/ 

/ 











/ 

/ 






■\ 













) 













1 













J 


























100 

s 

<- 

50 6 4 

uperheatF Moistu 

.re,pe 

8 12 
rcent 


Fig. 50. Correction Factors for Rotation Loss. 


It should be noted that the amount of rotation loss depends on 
the dimensions of the wheel and not on the weight of steam 
flowing. Hence it is relatively large for machines of small output 
or with large diameter wheels. If two rows of buckets are 
placed on one disk, Ld is not increased. Hence the total rota- 



















































186 


Thermodynamics, Abridged 


tion loss is not doubled. Rotation loss is reduced when the 
steam is initially superheated. It is very great where buckets 
revolve in very wet steam. For wheels larger than 36 in., Ld may 
be 5 per cent, greater than is indicated by Equation (80). 

Prob. 281. A wheel of 1.146 ft. dia. revolves at 10 000 
r.p.m. in dry steam at 1 lb. pressure. Buckets are 1.3 in. high. 
Find Ld and L b . (Ans., 103 and 560 B.t.u. per hr.) 

Prob. 282. The wheel in Prob. 281 revolves in a chamber in 
which the pressure is 1 lb. The nozzle inlet pressure was 150 lb. 
(dry steam) and the nozzle efficiency 0.93. What is the dryness 
at the nozzle outlet? (Ans., 0.80.) If the work at the buckets 
is 125 B.t.u., find Qh . (Ans., 1069.) What then is x *? (Ans., 
0.966.) What is the mean dryness in the wheel chamber? 
(Ans., 0.883.) What is the rotation loss correction? (Ans., 
1.765.) What is the corrected rotation loss? (Ans., 1170 B.t.u. 
per hr.) 

142. Efficiency and Steam Rate. The bearing friction and 
shaft packing losses of the turbine may be kept down to 1 per 
cent, of the shaft output. Radiation losses are small. Losses 
incurred in speed reduction are perhaps properly chargeable to 
the turbine in comparing its performance with that of a re¬ 
ciprocating engine. Efficiencies of reduction gears range up 
to 0.98: hydraulic transmissions and electric drive are tech¬ 
nically less efficient. The work at the buckets corresponds 
with the Ihp. of an engine, but cannot be measured directly. 
Turbine capacities and economics are therefore expressed on the 
basis of brake hp. or shaft hp. The ratio of shaft hp. to bucket 
hp. may be called the mechanical efficiency, although the 
meaning of this term is then not quite the same as with steam 
engines. This usage lumps the rotation loss with the mechanical 
friction loss. 

Let W = work at buckets per lb. of steam, in B.t.u., w = lb. 
of steam per hr. Then bucket hp. = Ww -S- 2545. Let L = 
total rotation loss, B.t.u. per hr. Then shaft input = Ww — L 
B.t.u. per hr. Let 1 — l = proportion of shaft input lost in 


Fluid Flow and the Steam Turbine 


187 


mechanical friction (say 1 per cent.). Then l(Ww — L) = shaft 
output, B.t.u. per hr., or l{Ww — L)/2545 = shaft hp. and the 
mechanical efficiency, as above defined, is 


l(Ww - L) -r- Ww = 


2545 X shaft hp. 
Ww 


e M- 


The efficiency from steam to work at the shaft is e — e z X e N 
X e k X e M . What corresponds with the “ relative efficiency ” 
of the steam engine (Art. 93) 


is the product e k X e N . This 
may be higher than in steam 
engines, and is chiefly deter¬ 
mined by the value of e k . If 
(Fig. 46) the steam and feed 
water conditions are Q a and 
h, the weight of steam per 
brake hp.-hr. is as in Art. 
93, wl hp. = 2545 ^ e(Q a - 
lib ) lb. Fig. 51 gives ideal 
steam rates for the complete 
expansion cycle (Arts. 80, 
84). Results from actual 
trials are given below, Art. 
148. 


7 yn~. 

-rNon Condensing lurbines- 
-16'lb. Absolute—|— 



80 ZOO ZZO Z40 
Absolute Initial Steam Pressure Ib.persq.in. 

Fig. 51. Steam Rates for Ideal Tur¬ 
bine Cycles. 


Prob. 283. The turbine gives 125 B.t.u. work at buckets per 
lb. of steam, and the rotation loss is 1170 B.t.u. per hr. Me¬ 
chanical friction is 1 per cent. The shaft hp. is 100. Find 
weight of steam per hr. (Ans., 2053 lb.) 

Prob. 284. In Prob. 283, what is the “mechanical efficiency , ’? 
(Ans., 0.985.) If the initial steam was at 150 lb. pressure, and 
dry, and the back pressure was 1 lb., with e N = 0.93, find e I} 
and e. (Ans., 0.28, 0.42, 0.108.) What is the “ relative 
efficiency ”? (Ans., 0.39.) What weight of steam is used per 
brake hp.-hr.? (Ans., 20.53 lb.) Check the last result from 
Fig. 51. 






































188 


Thermodynamics, Abridged 


Multi-Stage Turbines 

143. Velocity Compounding. In Fig. 466, there is a consider¬ 
able exit velocity, cj, which carries away with it kinetic energy. 
This velocity should be kept small, or else means should be 

provided for the further util¬ 
ization of the wasted kinetic 
energy. The method is sug¬ 
gested in Fig. 52. The steam 
leaving the buckets A is re¬ 
versed by stationary buckets 

B, projecting inward from 
the casing, and used again 
in a second set of buckets 

C. To decrease rotation loss 
(Art. 141), these additional 
buckets are mounted on the 
original disk. 

In Fig. 47, the triangles 
abc, chj duplicate those of 
Fig. 466. Now lay off jm 
parallel with ba, implying equality of angles a for both rows 
of buckets. The length of jm is that of cj, reduced by bucket 
friction as in Fig. 48. Lay off mn, parallel and equal to be. 
(Both rows of buckets move in the same direction at the same 
speed.) Draw jn and draw np = jn so that the relative an¬ 
gles jut and put are equal. (This equality does not always 
exist.) Lay off nq, less than np, the ratio of the two being given 
by Fig. 48. Lay off qr parallel and equal to mn. Then the 
additional work gained by the second row of moving buckets, 
at no additional cost for heat, is 



Fig. 52. Arrangement of Buckets in 
a Velocity-Compounded Turbine. 


w 2 = 


mn(mt + ns) 
25 028 


B.t.u. per lb. of steam. 


(81) 


If the point r lies to the left of a vertical line through n, the 
component ns is negative. 













Fluid Flow and the Steam Turbine 


189 


Such velocity compounding increases work and efficiency. 
In the present illustration, only two stages are used. This is, in 
fact, all that it would be profitable to use under the velocity 
relations shown by Fig. 47 to be existing. With lower values 
of u, the desirable number of stages increases. By velocity 
compounding, we may increase the kinetic efficiency at a given 
value of u, or may maintain some desired efficiency without 
using high values of u. Bucket friction establishes a limit to 
this procedure. 

Prob. 285. In Prob. 275, a second stage is added as described 
in Art. 143. What is the additional work obtained per lb. of 
steam? (Ans., 35 B.t.u., or 28 per cent.) 

144. Pressure-Compounded Impulse Turbine (Curtis Type). 
It is the ratio of bucket speed to steam speed which determines 



Fig. 53. Arrangement of Nozzles and Buckets in a Two-Stage Curtis Turbine. 




































190 


Thermodynamics, Abridged 


kinetic efficiency, rather than the bucket speed itself. Velocity 
compounding, as just described, is equivalent to increase of 
bucket speed but is limited by bucket friction. Pressure 
compounding attacks the problem from another angle, by 
decreasing the steam speed. This is done by expanding in 
stages. One set of nozzles expands the steam part way down 
to condenser pressure. If half the “ heat drop ” ab, Fig. 46, 



is realized in this stage (or set of nozzles), the outlet velocity 
resulting will be that in a single-expansion nozzle divided by V2 
(Equation (74)). This velocity is then used in a wheel which 
may be either simple or velocity-compounded. Leaving this 
wheel, the steam expands down to condenser pressure through 
the second set of nozzles and then strikes a second wheel. Fig. 53 
suggests the arrangement with two pressure stages (sets of 
nozzles) and two-stage velocity compounding in each pressure 







Fluid Flow and the Steam Turbine 


191 


stage. The number of pressure stages should increase as the 
peripheral speed, u, decreases, in order to maintain an advan¬ 
tageous ratio between u and the steam velocity. 

Assume steam at 150 lb. pressure, dry, with a condenser 
pressure of 1 lb. and a nozzle efficiency of 0.93. Locate the 
initial state, a, on the total heat-entropy diagram, as in Fig. 54. 
Draw ab vertically, locating b. Read off Q a = 1194, Q b = 877. 



Fig. 55. Two-Stage Velocity Diagram, Pressure-Compounded Turbine. 

Assume two pressure stages: therefore bisect ab, determining 
p c = 17 lb., Q c = 1035.5. The first stage nozzles will expand 
from 150 to 17 lb. pressure and the velocity resulting is 224 
Vo. 93(1194 - 1035.5) = 2720. Assume u = 400, k = 20° and 
equal relative angles. Construct the velocity diagram for two 
velocity stages as in Fig. 55, following Fig. 47. The work per 
lb. of steam in the first pressure stage is 
be 

2 5 o28 ( b 9 + ck + mt + ns) 

= (2550 + 1230 + 1095 + 215) = 81 B.t.u. 

Zo UZo 

The heat in the steam entering the second pressure stage is 
(Fig. 54) Q e = Qa — 81 = 1113. This determines the point e on 











192 


Thermodynamics, Abridged 


the line of 17 lb. pressure. Adiabatic expansion in the second 
pressure stage is represented by the vertical line ef, determining 
Qf = 941 B.t.u. Then the heat to be converted into velocity in 
this second pressure stage is Q e — Qf = 1113 — 941 = 172 B.t.u. 
In the first stage it was 1194 — 1035.5 = 158.5 B.t.u. The 
second stage will therefore have a higher initial velocity and 
will give more work per lb. of steam. 

It is usually desirable that the stage conversions be equal: 
i.e., that Q a — Q c = Q e — Qf. This leads to similar blading in 
the two stages. The conversions are unequal in this case 
because of reheat, i.e., the reconversion of velocity not utilized, 
into heat. To allow for the influence of reheat, we may arbi¬ 
trarily increase the drop in the first stage. This implies lowering 
the discharge pressure of the nozzles in that stage. An approxi¬ 
mation to the desired condition is obtained by starting with a 
heat drop which is a mean between Q a — Q c and Q e — Qf. 

Prob. 286. What is the mean value of the two heat drops in 
Art. 144? (Ans., 165.3.) If this drop is used for the first stage, 
what is the value of QJ (Ans., 1028.7.) What is then the 



^ 0 3 6 9 12 15 18 21 24 27 30 


No. of Stage 


Fig. 56. Steam Velocity, Reaction Turbine. 


value of Pc? (15.7.) Of the outlet velocity? (Ans., 2780.) 
Assume that the use of this velocity in the first stage velocity 
diagram, Fig. 55, gives work equal to 84 B.t.u. per lb. of steam. 
What then is the value of Q e ? (Ans., 1110 B.t.u.) Of p c ? 
(Ans., 15.7.) Of Qf ? (Ans., 944.7 B.t.u.) Of the heat drop 
in the second stage? (Ans., 165.3.) Of the work in the second 
stage? (Ans., 84 B.t.u.) What is the total work at the buckets 



















































Fluid Flow and the Steam Turbine 


193 


of the whole turbine? (Ans., 168 B.t.u. per lb. of steam.) What 
is the gain over the machine of Prob. 275? (Ans., 43 B.t.u. or 
34 per cent.) Note that this gain is secured even though u has 
been decreased. 

145. Reaction Turbine. The impact due to the velocity cj, 
Fig. 466, is a reactive impact. The steam is flowing away 
from the bucket. This reaction is less than the impulse produced 
by the inlet velocity ab. Under ordinary conditions, it must 
be less. If, however, the buckets are so shaped that some 
expansion (lowering, of pressure) occurs in them, there may be 
an increase of velocity instead of a decrease, from ac to ch, in 
spite of bucket friction. The reaction work may then equal or 
even exceed the impulse work. 


a 



Fig. 57. Velocity Diagram, Reaction Turbine. 

In Fig. 57, assume that the relative exit angle gcd is equal 
to the absolute inlet angle, a. Assume also that there is 
expansion in the buckets, so great that the relative exit velocity 
without friction would be cd > ab. Assume further that the 
amount of expansion is such that this velocity, reduced by friction 
as in Fig. 48, is ch — ab. Transpose the triangle chj to the posi¬ 
tion ah'j'. Here it will be symmetrically opposed to the tri¬ 
angle abc. 








194 


Thermodynamics, Abridged 


The work per lb. of steam, following Equation (79), is 

&0 ,, , v _ be X ch' 

25 028 bg + 9< ^ 25 028 B-t ' u ' 

About g as a center describe a circular arc bl. Draw cf vertically. 
Then, from geometry, be X ch' = cf 2 and 

W = 25028 = ( 15875) Bt-U ’ ^ 

This modified system of velocities is approximated in turbines 
of the reaction (Parsons) type. There is a further character¬ 
istic of reaction turbines. Since expansion is to occur in the 
blades, no nozzles are necessary. The steam entering the 
first row of blades is simply at steam pipe velocity, around 100 
to 200 ft. per sec. Hence peripheral speeds (which should be 
about half the steam speed) may be very low without sacrifice 
of efficiency. Each row of blades may be regarded as a set of 
nozzles designed for a very small heat drop. The reaction 
turbine is essentially a low speed turbine. It contains rows of 
moving buckets alternated with rows of stationary reversing 
buckets (like those of Fig. 47), projecting inward from the casing. 

Prob. 287. In Fig. 57, ab = 224, be = 100, tan a — 0.5. 
Find W. (Ans., 1.2 B.t.u.) 

146. Steam Velocities in Reaction Turbines. With low steam 
velocities, the work done in a row of buckets is necessarily very 
small. This is suggested by Prob. 287. Hence the number of 
rows is very large. If the average work per row is 5 B.t.u., and 
the total bucket work of the turbine 150 B.t.u., 30 stages or 
rows of buckets are required. 

Fig. 57 shows that the absolute exit velocity cj is less than 
the absolute entrance velocity ab. In passing to the second row 
of moving buckets, however, further expansion occurs in the 
fixed blades: so that the absolute entrance velocity of the 
second row is made somewhat greater than ab. Hence (if 
the value of u is the same for both rows) a greater amount of 
work will be done in the second row than in the first. 




Fluid Flow and the Steam Turbine 


195 


A reaction turbine, unlike one of the impulse type, carries its 
wheel chamber constantly full of steam. Its capacity is there¬ 
fore determined by the area of the annulus comprising the 
buckets, as well as by the steam conditions. Tip clearances 
must be low. This is not necessary in impulse turbines. 



Fig. 58. Parsons Reaction Type Steam Turbine. 


Fig. 58 shows a marine type of reaction turbine. Steam 
enters at E and moves to the left. The buckets are mounted on 
three wheels or drums, of increasing diameter toward the exhaust 
end. This increase is made in order to avoid the use of very 
Jong buckets (long radially) toward the low pressure end. The 
valve V is employed to admit high-pressure steam to the lower 
pressure stages. This is done when overloads are encountered, 































































196 


Thermodynamics, Abridged 


but involves some loss of efficiency, there being an impaired 
relation of velocities. The three short drums to the right of the 
steam inlet E may be provided with reversed buckets, then 
constituting an astern turbine. Economy is not necessary 
when moving astern: hence this system contains only a few 
stages. The reversing action is not considered in the velocity 
and pressure curves below the illustration. Means must be 
provided for directing the steam either to the right or to the 
left from E. Values of u for the various drums have ratios equal 
to those of drum diameters. A common successive ratio is a/2. 
The work done by the successive drums increases toward the 
exhaust end. With three drums, the relative amounts of work 
are often in the ratio 1:1: lj. 

Suppose the adiabatic drop to be as in Art. 144, 317 B.t.u. 
Bucket friction should be low because of the low steam speeds 
used, but high because of the large number of buckets. There 
is a leakage loss past the blade tips, which reduces efficiency. 
Allowing for leakage and rotation loss (Art. 141), the work at 
the buckets may be expected to be around 75 per cent, of the 
adiabatic heat drop in carefully designed machines. 

Assume.the probable work at the buckets to be in our case 
230 B.t.u., and that 60 B.t.u. are to be derived from a first drum. 
How many stages (rows of buckets) shall that drum have? 

Assume u— 100, a = 26°, initial steam velocity 257 ft. per 
sec. By the method of Fig. 57, the work done in the first stage 
is 1.45 B.t.u. It will be greater in later stages. The number of 
stages will therefore be less than 60 -v- 1.45, or less than 41. 

Assume a probable variation of steam velocities from beginning 
to end of drum as in Fig. 56. In general, this velocity increases 
more and more rapidly toward the low-pressure end, as here 
indicated. The curve may be represented by an equation in the 
form 

U= aX e bs , (83) 

where U — steam velocity, ft. per sec., 
s = the number of the stage, 

e — base of Napierian system of logarithms, and a and b 
are constants. 


Fluid Flow and the Steam Turbine 


197 


The curve is established when the steam velocities at two 
stages are known. We have assumed U = 257 when s = 1. 
Assume also that U = 364 when s = 30. Then 


Un 

U, 



log* 


364 
257 : 


296, 


6 


0 . 012 . 


U i = 257 = ae °- 012 , log e 257 = log e a + 0.012, a = 254. 
Then the equation of the whole curve is U = 254 X e°- 012s . 

Referring to Fig. 57, the work done at any stage is (Equation 
(82)), 

w _ cf = ; W - eg 2 

25 028 25 028 25 028 

(£7 cos a) 2 — (U cos « — u ) 2 2uU cos o; — w 2 

25 028 = 25 028 * 


If U = velocity at any stage, U\ = velocity at first stage, 
c = J7 i/m, then 


Then 


77 = = c &(8-1) and U — cuX c 6(8-1) . 

C/i c 6 


2c^ 2 cos a X c 6( ® _1) — u 2 
25 028 


^ 2 

25 028 


(2c cos a X c b(s_1) — 1) 


The work of the whole drum is 


2JF = 


u* 


25 028 


^2c cos a j|* e H8 V ds — ds'j 
| 2c cos a ^ 


^(s-D _ i 


25 028 

For the conditions assumed, 


)-s). 


(84) 


10 000 f 
ZW ~ 25 028 I 


/ . 0 . 012 ( 3 - 1 ) _ l \ 

2 X 2.57 X 0.8988 ( —-J 


To find the value of g 0 012 ^ -1 ), put this quantity equal to x. Then 
log e x = 0.012(5 — 1), 

zw - 0.4(4.62 

4-6 ^q~ 1} = 150 + s = 385 Or - 1), 
log (535 + s) - 0.00522s = 2.581. 
















198 


Thermodynamics, Abridged 


By trial and error, s is found to be between 33 and 34. A whole 
number must be chosen. Select 34. Then in Equation (84), 


/ £>0.396 _ 1 \ 

= 0.4(186.3 - 34) = 60.92 B.t.u., 


instead of 60, as originally desired. The steam velocity at the 
34th stage may now be computed. It will be the initial steam 
velocity of the next succeeding drum. 



The whole rotor of a reaction turbine is subject to an end 
thrust due to the static pressure of the steam toward the low- 
pressure end. This is often counterbalanced by “ dummy 
pistons ” on which the areas are equivalent to the exposed 
drum areas while steam pressures are in the opposite direction. 
End thrust can also be eliminated, as in Fig. 59, by supplying 
steam near the middle of the length and allowing it to flow both 
ways. This illustration shows another common feature, the 
use of an impulse wheel (one set of nozzles, two velocity stages) 
preceding the reaction elements. A large proportion of the 
heat is converted into work at the impulse wheel, at a point 
where the steam is fairly dry and the rotation loss (Art. 141) 
consequently small. Hence the reaction part of the turbine is 
reduced in size and contains fewer stages. Furthermore, tip 

































Fluid Flow and the Steam Turbine 


199 


leakage (which is most objectionable at the high-pressure end 
of the reaction turbine) is limited to the low-pressure reaction 
stages. 

Prob. 288. Check the following values of Art. 146: 1.45, 
0.012, 254. What would be the work of the drum if 33 stages 
were used? (Ans., 58.9 B.t.u.) What is the steam velocity at 
the 34th stage? (Ans., 382 ft. per sec.) 

Turbine Economy 

147. Low Pressure Turbines. The turbine works with com¬ 
plete expansion while the engine does not. The former can do a 
great deal with steam rejected by an engine. Thus, in Fig. 26, 
suppose steam at 5 lb. pressure, 0.80 dry, to be available. Its 
heat content is 933 B.t.u. Expanded adiabatically to 1 lb. 
pressure, its content becomes 849 B.t.u. The ideal work is then 
84 B.t.u. per lb. of steam, and the actual work might easily be 
60 per cent, of this. 

Prob. 289. A steam engine receives dry steam at 270 lb. 
pressure and expands it to 5 lb. pressure. A turbine then 
expands it down to | lb. pressure. Ideally, what proportion of 
the total power is derived from the turbine? What is the ideal 
steam rate for the combination? (Ans., the turbine gives 42 per 
cent, of the power of the engine. Steam rate, 6.6 lb. per Ihp.-hr.) 
What would have been the ideal rate of a turbine alone, using 
dry steam at 270 lb. pressure and expanding to J lb. back pres¬ 
sure? (Ans., 6.6 lb.) 

148. Tests of Steam Turbines. Following are ranges of steam 
rates from many trials: 


Lbs. Steam per Brake 

hp.-hr. 


Vacuum, in. 

Saturated steam: Simple impulse type.. 

. 14 to 

23 

25 

to 

28 

Pressure compounded. 

. 16 to 

20 

27 

to 

29 

Reaction type. 

.14 to 

15 

27 

to 

28 

Superheated steam: Simple impulse type.. 

.12| to 

15| 

27 

to 

28 

Pressure compounded. 

. 12* to 

16* 

28 

to 

291 

Reaction type. 

.11* to 

141 

27 

to 

29 

Many recent tests show improvement on these 

s figures, 

which 

in 

no 


involved very high superheat. 
14 




(Carves A,B,U,D) (Curve E only) 


200 


Thermodynamics, Abridged 


Normal atmospheric pressure is 14.696 lb. per sq. in. or 29.92 
in. of mercury. Vacuum is measured downward from atmos¬ 
pheric pressure. If the vacuum gauge reading is po in. of mercury 
the absolute pressure is (14.696/29.92) (29.92 — p 0 ) = 0.491 (29.92 
— p 0 ) lb. per sq. in. 



Fig. 60. Turbine Efficiency as Affected by Load. 


Fig. 60 shows the effect of variable load on turbine economy, 
the curves being similar to those for steam engines (Fig. 29). 
Particulars follow: 

A and B, 1050 hp. Westinghouse-Parsons machine, 165 lb. 
pressure, 28 in. vacuum. A , saturated steam; B, steam 
superheated 100°. 

C and D, 400 hp. Westinghouse turbine, 28 in. vacuum: 
C, saturated steam; D, steam superheated 90°. 

E, 9000 kw. 5-stage Curtis turbine, 215 lb. pressure, 125° 
superheat, 29 in. vacuum. 

Fig. 61 illustrates the effect of varying superheat. The steam 
rate tends to bear a straight line relation with the amount of 
superheat. The lines are, 








































Fluid Flow and the Steam Turbine 


201 


A, 2000 hp. Curtis turbine. 

B and C, average results: B for 26 in. vacuum, C for 28 in. 
vacuum. 

D and E, 400 hp. Westinghouse machine, 28 in. vacuum. D , 
1.3 full load; E, full load. 


U_l 



Fig. 62 shows a straight line relation between vacuum and 
economy. A is a 2000 hp. Curtis machine; B, a 4750 kw. turbine; 
C and D are 1500 kw.: D is at § load, C at full load. 

Prob. 290. A recent turbine is guaranteed to use not over 
13 lb. of steam per kw.-hr. with steam at 300 lb. pressure, super¬ 
heated 300° and at 28.9 in. vacuum. Are there any rates shown 
in Art. 148 or Figs. 60-62 indicating that this is possible? What 
is the ideal steam rate per hp.-hr. under these steam conditions? 
(Ans., 5.44 lb.) Per kw.-hr.? (Ans., 7.3 lb.) If the efficiency 
from shaft to switchboard is 0.93, the guarantee implies what 
steam rate per brake hp.-hr.? (Ans., 9 lb.) If the mechanical 
efficiency is 0.95, what is the relative efficiency? (Ans., 0.64.) 












































202 


Thermodynamics, Abridged 


Prob. 291. Find the vacuum in in. of mercury at normal 
atmospheric pressure when the back pressure is 1 lb. absolute. 
(Ans., 27.88 in.) 

Prob. 292. The barometer reads 29.4 in. The vacuum gauge 
reads 27 in. What is the absolute back pressure? (Ans., 1.18 lb.) 

Prob. 293. In Fig. 60, what percentage reduction in steam 
rate results at normal load by substituting for saturated steam, 
steam superheated 90° or 100°? (Ans., 8.6 and 9.5 per cent.) 

Prob. 294. What is the weight of steam used per hour at 100 
per cent, overload by turbine B, Fig. 60? (Ans., 29 000 lb.) 



Vacuum In. 

0.5 0.75 1 1.25 1.5 175 2 

Exhaust Pressure-Lb.perSci.ln. 

Fig. 62. Turbine Efficiency as Affected by Vacuum. 


Prob. 295. In Fig. 60, which is the most economical turbine 
at normal load? At loads varying from 60 to 140 per cent, of 
rated capacity? 

Prob. 296. In Fig. 61, state for all cases the percentage 
saving in weight of steam due to superheating 100°. (Range 
is from 7 to 11 per cent.) 

Prob. 297. In Fig. 61, with dry steam, what is the percentage 
reduction in steam rate by increasing the vacuum from 26 to 
28 in.? What is it when the steam is superheated 100°? 160°? 

(Ans., 8, 7, 6| per cent.) 

Prob. 298. State in words the significance of the crossing of 
curves C and D, Fig. 62. 




















CHAPTER IX 


INTERNAL COMBUSTION ENGINES 

Four-Cycle Engines with Explosive Charge 
149. The Otto Cycle. Fig. 63 is an indicator diagram from a 
gas engine. Note that it represents four successive strokes. 
At the point c, the piston is at the left-hand end of its stroke, 
the exhaust valve is closed and the inlet valve is open. The 
piston moves one full stroke to the right, drawing in a charge of 



combustible mixture. This operation is represented by the 
line cd, along which the pressure is slightly below that of the 
atmosphere. 

At d, the inlet valve closes. The piston now makes a stroke 
to the left, compressing the combustible mixture along de • • •. 
At or near the end of this stroke, the charge is ignited by appro¬ 
priate means, and there is a rapid combustion accompanied by a 
sudden rise of pressure. The piston then moves to the right 
under the expansive action (falling pressure, increasing volume) 
of the heated gas. At or near the end of this stroke, the exhaust 
valve opens and the burnt mixture is discharged (line be) at a 
pressure slightly above that of the atmosphere. Fig. 65 shows 
the engine. 


203 








204 


Thermodynamics, Abridged 


150. Ideal Cycle. This operation is idealized in Fig. 64, 
diagram abed. Here the counter-clockwise (power-coNSUMiNG) 
area due to intake and exhaust is disregarded, the combustion 
is assumed to be perfectly instantaneous and compression and 
expansion are taken to be adiabatic. The cycle is therefore 
bounded by two adiabatics and two lines of constant volume. 



Fig. 64. Ideal Otto Cycle. 


Remembering that efficiency is (heat absorbed — heat re¬ 
jected) -t- heat absorbed, and that there is neither absorption 
nor rejection of heat along an adiabatic, 

Hbc H da Hda .. wl(Td T 0 ) Td — T a 

H hc “ Ihc ~ wl(T c - T b ) = ~ TV^Tb' 


e = 










































Internal Combustion Engines 


205 


the specific heat at constant volume being denoted by l and the 
weight of mixture by w. Now the paths cd and ba are isodiabatics 
(Art. 31). Hence T d /T a = TjT b , 



if the mixture is regarded as a perfect gas. In these expressions, 
C = COMPRESSION RATIO = Pb/pa] C = CLEARANCE = F&/(F a 
— V b ). One of the expressions looks like that previously given 
(Art. 34) for the efficiency of the Carnot cycle. It has a different 
meaning: T b and T a are not the extreme temperatures of the 
cycle. The efficiency of the Otto cycle is less than that of the 
Carnot cycle. It is much greater than that of the Rankine 
cycle, used for the steam engine and turbine: mainly because of 
the very great temperature range employed in gas engines. 
Clearance in gas engines is made to suit the compression desired. 
It is not to be viewed as a “ necessary evil,” as it is in steam 
engines. 

Prob. 299. Given C = 10, find efficiency and clearance, using 
y = 1.4. (Ans., e = 0.48, c = 0.24.) 

151. Mean Effective Pressure. Still considering the sub¬ 
stance to be a perfect gas, 


2IF 

Vm ~ 144(F„ - Vb) 


PcVc — PdVd — PbVb + PaV a 
(y - 1) (Fa - Vb) 


(see Art. 28) 


Vc(Pc — Pb) + V a(p a — Pd) 

(y-l)(V a -Vb) 


But F c /(Fa — Vb) = c: while 

V h /l \ l 'y 

(Fa - Vb)IV a = 1 - pT “ 1 - (c) = 1 - C ~ il,V) - 

Then, remembering that p a = Pb/C and pa = pdC, 








206 


Thermodynamics, Abridged 


_ c(p c - p t) Vb - Vc 

Vm~ y _ 1 + C(1 _ C-U/ri) 

<«> 

A more useful expression is 

7782 H _ 5.mwl(T c - Tt)e 
Vm ~ 144(Fa _ y b) ~ V a -V b 

This has a special meaning. The quantity wl(T c — T b ) repre¬ 
sents the heat available in the cylinder for raising the tempera¬ 
ture of w lb. of mixture, at constant volume, from T b to T c . 



It is the heat value of the mixture present. All of that heat, 
under ideal conditions, is used to increase the temperature 
along be. Now consider the hitherto ignored intake stroke, ea, 
Fig. 64. At the beginning of this stroke, the volume of exhaust 
gas in the cylinder was V e . At the end of the intake stroke, the 
volume of exhaust gas and fresh mixture was F 0 . Hence the 
volume of fresh mixture admitted, and present during com¬ 
bustion, was V a — V e — V a — Vb- Then the heat value 


































Internal Combustion Engines 


207 


OF 1 CU. FT. OF MIXTURE SUPPLIED was wl(T c — T b ) -5- (F 0 — V b ). 
Call this B. Then 

p m = 5A0SBe, lb. per sq. in. (87) 

Here e = efficiency, from Equation (85). 

Prob. 300. In Prob. 299, find p m from Equation (86) if 
p a = 14.696, p c = 346.96. (Ans., 58 lb.) What heat value 
per cu. ft. of mixture does this imply? (Ans., 22 B.t.u.) 

152. Variable Specific Heat. At the high temperatures pre¬ 
vailing in the cylinder of an internal combustion engine, the 
specific heats are highly variable, although the gases are not 
so imperfect as to depart from Equation (12). Some refinement 
is therefore necessary over the foregoing. 

Suppose l = a + bT, where a and b are constants. Then 

H bc = f C IdT = PW + bTdT 

b *Jb b 

= a(T c - T h ) + \ (T > - 2V). 

Similarly, H ai = a( T d - T a ) + (6/2) (7 1 V - 27), and 

H bc Had Had 
e= lhT~ = 1 _ hZ 

a{T d - T a ) + x (27 — T a 2 ) 

= 1 - r -• ( 88 ) 

a(T c - T b ) + g (27 - 2ft 2 ) 

Under these conditions, adiabatics are not isodiabatics, for the 
equation of the adiabatic is not pV y = const. On the contrary, 
if we may write k = o + bT, R /778 =R' = k—l=o — a, 
the equation of the adiabatic (with variable specific heats) is 
thus derived: 

In an adiabatic process, the external work done equals the 
loss of internal energy: or T = - W (Art. 28). Applying this 
to a very brief process, during which the pressure is substantially 
constant, IdT = - Pdv /778 for 1 lb. weight. Writing l = a 
+ bT, P = RT/v, this may be reduced to 




208 


Thermodynamics, Abridged 


adT R'dv 

— +hd T= • 

Now differentiate the perfect gas equation, Pv = RT. This 
gives Pdv + vdP = RdT. Dividing by Pv, 

dv dP _ dT 

7 + T~ T m 

Substituting this value in the first term of the preceding equation, 
we obtain 

(a+ R')j+ a^+bdT = 0. 

Integrating, (a + R') log e v + a log e P + bT = const. 

- loge v + loge P + - T = const., 

0/ CL 

Pv 0,a e bTla = const., (89) 

where e is the base of the Napierian system of logarithms. 



Fig. 66. Otto Cycle Efficiencies, Constant and Variable Specific Heats. 










































































































Internal Combustion Engines 


209 


In the ideal case, the mixture is considered to be pure air. 
It is, in fact, mostly air. Values of a, b and o are not exactly 
known: but approximately, a = 0.1801, b = 0.0000283, o = 
0.2511. Fig. 64 shows by dotted lines the resulting cycle, to 
scale, though not for these particular values of the constants. 
In both diagrams, p a = 14.696, T a = 500, C = 10, and the 
heat evolved during combustion is 700 B.t.u. per lb. of mixture. 
The chief difference is with respect to the rise of temperature 
along be. With l = 0.17, this is 700 -i- 0.17 = 4120°. 

Fig. 66 compares efficiencies of the two cycles, the broad band 
of results from Equation (88) covering the whole range of pro¬ 
posed values of a, b and o. The heavy dotted line within this 
band represents an efficiency which is 81.8 per cent, of that by 
Equation (85). This line may be regarded as representing results 
of Equation (88) as closely as they can be known at present. 
In other words, the ideal efficiency is 81.8 per cent, of that 
given by the simple (constant specific heat) expression of Equa¬ 
tion (85). 

Prob. 301. In Art. 152, show that the heat absorbed along 
be' is approximately 700 B.t.u. Take values of constants from 
Fig. 64. 

Prob. 302. From Equation (89), show that 


1 Th .l h -lT T \ " ~ ° l nrr 

log ‘ZV + a (n Ta) ~ a ° gc V a ’ 


O. Tub ms _ 0 a. Vb 

-l°g.jr + - ( n- fa) - a Iog. p> . 


Prob. 303. Does Charles’ law apply when the specific heats 
are variable as in Art. 152? 

Prob. 304. What is the probable true ideal efficiency when 
C = 10? (Ans., 0.397.) 

153. Variations of Ideal Efficiency with Load: Governing. 

When the load on an engine decreases, the supply of heat should 
be decreased. This is accomplished in a constant speed engine 
in one of four ways: 




210 


Thermodynamics, Abridged 


(1) Vary the time of ignition. This simply throws away heat 

not needed. 

(2) Omit drawing in a charge for one or more strokes (“ hit 

or miss ”). This leads to irregular speed and is applic¬ 
able only to small engines. 

(3) Dilute the charge with an excess of air. 

(4) Decrease the weight of charge drawn in. 

Methods (3) and (4) are important. We have seen that 
efficiency depends on compression (Art. 150, Equations (85), 
Fig. 66) : strictly, on the compression ratio, C. Compression 
is limited by the temperature attained at b, Fig. 64. We must 

Typical Ignition Temperatures, °F.* 

Ether, 750 Acetylene, 760-820 Benzene, 780 Alcohol, 950 
Liquid Fuels, 950-1020 Equal Parts H and O, 957 Benzol, 970 
Ethylene, 1005-1015 Hydrogen, 1075-1100 

Illuminating Gas, 1100 Methane, 1200-1240 


not reach the ignition temperature by compression, else pre¬ 
ignition (self-ignition) will occur. For each possible mixture, 
there is a limiting permissible compression and hence a limiting 
efficiency. We should work as close to this compression as 
we can. In fact, without compression it would be difficult or 
impossible to ignite some fuels at all. The accompanying table 
lists the more common mixtures in the approximate order of 
allowable compressions, and therefore in the approximate order 
of efficiencies. (The values of the last column are discussed 
later.) 

Now if we dilute the charge with excess of air, we raise its 
ignition temperature (and also make ignition occur more slowly). 
Hence we are operating at an efficiency below the best possible 
efficiency. 

If we decrease the weight of charge, we also dilute it. This 
would occur, if for no other reason, because the charge is mixed 
with an approximately constant quantity of exhaust gas left 

* Except as noted, in air at normal pressure. 




Internal Combustion Engines 


211 


in the clearance space. Here also we are operating, when at 
reduced load, at less than maximum possible efficiency. 


Usual Compression and Mean Effective Pressures for Various Fuels* 


Fuel and Engine 

Compression 

Pressures 

Pm' 

Usual Range 

Average 

Kerosene in small hot bulb engines. 

45- 90 

75 

50- 80 

Gasoline, automobile. 

60-110 

80 

70- 90 

Kerosene, carburettor engine. 

60-100 

80 

50- 75 

Fuel oil, hot bulb engine. 

— 

60 

50- 85 

Gasoline, stationary and marine engines. 

75-120 

85 

80- 95 

Heavy fuel oils. 

75-135 

— 

50- 70 

Rich gas mixtures. 

70- 90 

— 

70- 85 

Lean gas mixtures. 

90-135 

— 

65- 80 

Natural gas, large engines. 

90-175 

135 

80-110 

Coal gas, small engines. 

90-135 

115 

60-100 

Carburetted water gas, small engines. 

90-120 

105 

60-100 

Producer gas, general. 

115-175 

145 

55- 90 

Coke oven gas, large engines. 

120-150 

135 

80-110 

Blast furnace gas, very large engines. 

135-205 

170 

80-120 

Alcohol, carburettor engine. 

135-225 

165 

130 


In either case, the efficiency at light loads is less than it should 
be, and (actually) much less than at full load. Methods (3) 
and (4) for governing are generally (and unavoidably) used 
simultaneously. When the mixture is diluted by throttling the 
gas, the intake suction is increased and the pressure at a, Fig. 64, 
is reduced. The weight is therefore reduced. When the weight 
of charge is decreased, the increased intake suction modifies the 
proportions of the charge. 

An important defect in gas engine governing arises from the 
fact that the governing action (by methods 2, 3 or 4) is exerted 
one full revolution before its effect is felt. 

Governing by dilution of charge naturally decreases the 
B.t.u. per cu. ft. of mixture and hence the value of p m (Art. 151). 
Governing by decreasing weight of charge reduces the B.t.u. 
per cu. ft. of displacement and hence the value of p m . 

Prob. 305. What is the best ideal efficiency to be expected 
with producer gas, if the pressure at a, Fig. 64, is 14.6 lb.? 
(Ans., 0.417.) 

* Mostly from Lucke. 

























212 


Thermodynamics, Abridged 


Prob. 306. State an average value of C for alcohol, if p a 
= 14.35. (Ans., Ilf.) 

Prob. 307. In an engine governed by varying the weight of 
charge, how would the efficiency vary with the load if the clear¬ 
ance were perfectly adjustable? 

154. The Actual Engine. In the actual engine, the intake 
pressure is -somewhat below atmospheric and p a = 8§ to 14f 
The compression is not precisely adiabatic. It may be repre¬ 
sented by pv n = const., with n between 1.25 and 1.35. Ignition 
is not instantaneous and the combustion line has a perceptible 
slope to the right as it rises. This is accentuated at high piston 
speeds or with slow ignition (improper mixture) conditions. 
The speed of flame propagation is of the order of 7 to 20 ft. per 
sec. at the pressures existing and with correct mixtures. The 
maximum temperature and pressure are much less than those 
computed as for Fig. 64 by either method. This is partly because 
the volume increases during combustion and partly because of 
the cooling which occurs. The maximum temperature is rarely 
much over 3000° F. The expansion curve is a polytropic with 
n between 1.3 and 1.5. The temperature at the point when the 
exhaust valve opens may be anywhere from 1200° to 2000°. 
The exhaust valve opens considerably before the end of the 
stroke. The average exhaust pressure is 16 or 17 lb. The 
temperature in the exhaust pipe is 700° to 1000°. 

155. Diagram Factor and Volumetric Efficiency. The actual 

indicated thermal efficiency is usually from 40 (occasionally down 
to 30) to 55 per cent, of that given by Equation (85), or 49 to 67 
per cent, of that by Equation (88), for the compression used. 
The “ relative efficiency ” (Art. 93) is therefore about the same 
as for steam engines: but its variations have not been as thor¬ 
oughly studied. The lower values of relative efficiency are 
obtained with engines employing carburettors to vaporize the 
fuel. This percentage which the actual indicated thermal 
efficiency bears to the ideal efficiency is also called the diagram 
factor. 


Internal Combustion Engines 


213 


The volumetric efficiency is the ratio of the volume of 
gas drawn in (measured at normal barometer and 62°) to the 
corresponding piston displacement. In discussing the ideal 
cycle, Art. 151, it was considered to be 1.0. In practice, it is 
decreased by the excess of the exhaust pressure over 14.696, by 
the suction below that amount, and by any intake temperature 
above 62°. Values of the volumetric efficiency are e v = 0.50 
to 0.87. It is lowest in high speed engines and with automatic 
valves. It is particularly low in airplane engines when flying 
at high altitudes. Supercompression involves filling the 
cylinder by means of a special external pump: by which method 
e v may be greatly increased. 

The actual mean effective pressure is then to be obtained 
from a modification of Equation (87): 

pj = 5A03fBee v , (90) 

where B = B.t.u. per cu. ft. of mixture, / = diagram factor, 
e = ideal efficiency. Values of pj are given in the second table 
of Art. 153. The ratio of actual to ideal mean effective pressure 
(p m '/p m ) is about equal to the ratio of actual and ideal pressure 
rises, p c — Pb, Fig. 64. 

Prob. 308. What maximum mean effective pressure may be 
expected with blast furnace gas (60 B.t.u. per cu. ft. of mixture) 
in a large slow-running engine with mechanically operated valves? 
(Ans., 87 lb.) 

Prob. 309. An automobile engine gives 65 lb. mean effective 
pressure. It has average compression. Assume 100 B.t.u. per 
cu. ft. of mixture. Make an estimate of its volumetric efficiency. 
(Ans., 0.82.) 

Prob. 310. Compute B for blast furnace gas, from the follow¬ 
ing table. (Ans., 57.) 

Prob. 311. Assume gasoline to be represented by C 6 H h , and 
to contain 19 600 B.t.u. per lb. What is its heat value per cu. ft. 
of vapor at standard conditions (32°, 14.696 lb.)? Compute B. 
(Ans., 4710 and 96 B.t.u.) 


214 


Thermodynamics, Abridged 
Properties of Gas Fuels (Lucke) 





Composition by Volume 



B.t.U. 


CH4 

CeHs 

CO 

o 2 

h 2 

CO2 

n 2 

per 

Cu. Ft* 

Natural gas 

(Kansas). 

0.982 

0.001 

0.0025 

0.0025 




950.1 

Coke oven gas. 

0.343 

0.040 

0.0600 

0.0110 

0.420 

0.025 

0.101 

579.8 

Retort coal gas .... 

0.3135 

0.022 

0.0860 

0.0035 

0.525 

0.015 

0.035 

542.6 

Oil gas. 

0.525 

0.235 

0.0100 

0.0050 

0.185 

0.005 

— 

1196.5 

Blast furnace gas .. 

0.002 

— 

0.2861 

— 

0.027 

0.114 

0.571 

107.5 

Water gas. 

0.044 

0.001 

0.4485 

0.0050 

0.456 

0.045 

0.001 

329.9 

Producer gas (from 
hard coal). 

0.002 

— 

0.2610 

0.0020 

0.150 

0.053 

0.532 

134.7 


156. Mechanical Efficiency: Overload Capacity. If we follow 

steam engine practice, the mechanical efficiency is e M = bhp. 
-T- ihp., where ihp. refers to the net diagram of Fig. 63; i.e., to 

* Low value, i.e., heat evolved when the products of combustion are not 
condensed in the calorimeter: taken at 32° and normal barometer. Note: 
the values given in the last column are not those of B, Equation (90). That 
symbol stands, not for the heat value of the fuel, but for the heat value of 
1 cu. ft. of combustible mixture, formed by adding the proper amount of air 
to the fuel. The weight of chemically ideal mixture is computed by utilizing 
the law that combining proportions by volume are those of the number 
of molecules entering into the reaction. Thus CO and H 2 require 
half their volume of oxygen, C 2 H 4 three times its volume of oxygen, CH 4 
twice its volume and CeH 6 seven and one half times. From the total oxygen 
requirement thus computed deduct the oxygen in the fuel and multiply the 
remainder by 4.75 to obtain the volume of air which must be added to 1 cu. ft. 
of fuel to produce a perfect mixture. Thus, for natural gas: 

0.982 X 2 = 1.964 
(C 6 Hj) 0.001 X 7\ = 0.0075 
0.0025 X \ = 0.0013 

1.9728 

0.0025 


1.9703 X 4.75 = 9.37 

1.0 


950.1 -f- 10.37 = 92 B.t.u. per cu. ft. mixture 


At 62° and normal barometer, 


at 32° F. 


B 


= 92 


460 + 32 
X 460 + 62 


861 B.t.u. 


The mixture obtained in practice is never exactly right. Hence B is less 
than this. If the mixture is far from correct, combustion is slow. If very 
weak or very strong, the mixture may refuse to ignite at all. 

























Internal Combustion Engines 


215 


the clockwise cycle minus the negative loop. It is logical, how¬ 
ever, to regard the ihp. as the larger value due to considering the 
clockwise cycle alone, without deduction for the loop, because 
that loop is due to friction. The friction is of course fluid friction 
rather than mechanical friction. The loop consumes 5 to 10 
per cent, of the power of the engine. If we regard the clockwise 
diagram as representing the ihp., the mechanical efficiency ranges 
from 0.75 to 0.85, increasing for large engines and being higher 
for double-acting than for single-acting engines (Art. 7). Let 
A = gross ihp., C — loop ihp., D = brake hp.: then as now 
defined e M = D/A wdiereas as defined for a steam engine 

= DI<A - C). 

Equation (90) shows that p m ' varies with/, B, e and e v . Natur¬ 
ally, / and e v will be kept as high as possible at the rated load 
condition. The value of e depends on the compression alone, 
and that will be as high as the nature of the fuel will permit. 
The effort will be made to have the mixture proportions right. 
Hence all factors have maximum values at rated load. In a 
constant speed engine, there is consequently no way of increasing 
power beyond that for which the engine was built. Efficiency 
and output vary in the same direction, not at all as in a steam 
engine. Maximum efficiency occurs at maximum power. An 
Otto cycle engine has, strictly speaking, no overload capacity. 
A margin of power for emergencies can be provided only by 
rating the engine at a power less than that obtainable at maxi¬ 
mum efficiency (Art. 165). 

Prob. 312. The gross ihp. is 100, loop hp. = 8, brake hp. 
= 80. Find two values of e M . (Ans., 0.87 and 0.80.) Which 
is adopted in the text?, (Ans., 0.80.) 

Prob. 313. Sketch a curve of thermal efficiency against brake 
hp. Compare Fig. 71. 

Other Otto Cycle Engines 

157. Two-Cycle Engine. The action of the two-cycle engine 
is suggested (and somewhat exaggerated) in Fig. 67. The 


216 


Thermodynamics, Abridged 


exhaust opens at d, quite early in the expansion stroke. When 
the pressure has fallen to p e , the inlet opens to admit a charge 
of mixture under pressure. This flows in along efg. The 
exhaust does not close until point a is reached. Compression 
then begins. A separate pump or its equivalent is provided for 



delivering the mixture to the cylinder. Its indicator diagram ; 
shown dotted and with the V axis reversed, corresponds with 
the negative loop of Fig. 63: but the loop now consumes 7 to 12 
per cent, of the engine power and the mechanical efficiency as 
defined in Art. 156 is reduced. It is from 0.65 to 0.70, being 
high for large engines. 

Along efg, inlet and exhaust valves are both open. Hence 
some fuel will be lost. This is mitigated by scavenging, i.e., 






Internal Combustion Engines 


217 


by having a blast of pure air precede the injection of the charge: 
but it is never entirely eliminated. The two-cycle gas engine is 
about 20 per cent, less efficient than the four-cycle. If it were 
not for its larger fluid friction loss, it would give twice the power, 
because only two strokes, instead of four, are necessary to com¬ 
plete a cycle. (Exhaust and intake are crowded into a small 
portion of the expansion and compression strokes.) Actually, 
the gain in power is about 70 or 80 per cent. 

The necessary outside compression of mixture may be pro¬ 
vided by (a) using the idle end of an otherwise single-acting 
cylinder: ( b ) utilizing the crank case as a compression chamber: 
(c) separate pumps. Methods (a) and ( b ) make the parts of the 
engine inaccessible and complicate lubrication and cooling. A 
two-cycle engine may be valveless, the passages being slots in 
the cylinder wall, opened and closed by the piston. 

In the smaller and cheaper two-cycle engines, e v is particularly 
low. The ideal efficiency is that given by Equations (85) or (88). 
Diagram factors and mean effective pressures are about 0.8 those 
of four cycle engines. 

Prob. 314. State the probable actual indicated thermal and 
brake thermal efficiency, and mean effective pressure, of a large 
slow-running two-cycle engine using blast-furnace gas. See Art. 
155, Prob. 308, Prob. 310, Art. 156. (Ans., 0.25, 0.165, 63 lb.) 

158. Size and Power of Engine. Following Art 7, for a four¬ 
cycle single-acting cylinder (which yields power at one-fourth 
the rate of a steam cylinder) 

ihD _ JL d 2 V ' X ILK- _ IPulMK 

inp * “ 16 Vm A 33 000 264 000 

_ Pw/AS _ p m LAN /q-i\ 

“ 132 000 ” 66 000 * { ' 

For other types, use the following multipliers for the above: 

2 for four-cycle double-acting, 

2 for two-cycle single-acting, 

4 for two-cycle double-acting. 






218 


Thermodynamics, Abridged 


Most engines have a plurality of cylinders, all of which develop 
equal power. Double-acting engines are difficult to lubricate 
and cool. 

Prob. 315. In Prob. 314, find the diameter and stroke of an 
engine to develop 500 brake hp. at 600 ft. piston speed and 
110 r.p.m., if two-cycle, two-cylinder, single-acting. (Ans., 
23.6 by 32.7 in.) 

Prob. 316. The commercial formula for brake hp. of an 
automobile engine is nd 2 j 2.5, where n = no. of cylinders, d = dia. 
in inches. If this is based on 1000 ft. piston speed and e M = 0.75, 
for single-acting four-cycle engines, what mean effective pressure 
is assumed in the cylinders? (Ans., 90 lb.) 

159. Oil Engines. Engines employing carburettors to vapor¬ 
ize a volatile fuel are not to be classed as oil engines, although 
kerosene, and some few heavier oils, may be used in such engines 
if the carburettor is heated. An oil engine, properly so called, 
draws in pure air on its suction stroke and the fuel is injected 
separately. A two-cycle oil engine may therefore avoid the 
loss of fuel to the exhaust (Art. 157) which is the largest factor 
in producing low efficiencies of two cycle engines. 

The less volatile fuels are employed for oil engines. They can¬ 
not be ignited by electric spark. Hence such engines are pro¬ 
vided with an uncooled “ hot bulb,” “ hot tube ” or plate, 
which communicates with the clearance space. This is kept red 
hot. It is heated by a torch for starting. It ignites the com¬ 
bustible mixture only after the temperature of the air in such 
mixture has been somewhat increased by compression. 

Some oil engines inject the fuel during the suction stroke or 
the early part of the compression stroke. The whole charge 
is then heated by the compression before it actually ignites. 
The time of ignition is determined by the proportioning of the 
uncooled surface, the compression, etc., and is apt to be irregular 
at variable loads. In other engines, there is timed injection 
of the fuel, close to the end of the compression stroke. Ignition 
cannot then be premature. Further, the compression may 


Internal Combustion Engines 


219 


be carried very high and high efficiency secured. On the other 
hand, the fuel must be injected against a high resisting pressure. 
Heavy oils must be finely subdivided in order that ignition may 
occur. Hence the best method of delivering them to the cylinder 
is by a compressed air blast. The necessary air pressure for 
maximum subdivision of the oil particles is about twice the 
pressure in the cylinder (Art. 136). Hence with high com¬ 
pressions, very high air pressures and multi-stage air com¬ 
pression are necessary for feeding the fuel. 

Oil engines of this type are easily governed by varying the 
quantity of fuel injected. The diagram factor decreases some¬ 
what at light loads. Overloads lead to over-rich mixtures,, 
overheating and a smoky exhaust. Compression pressures 
(pb, Fig. 64) as high as 300 lb. are being used, but these lead to 
extremely high maximum pressures. 

For liquid fuels in general, the specific gravity is 

140 

5 “ 130 + 6’ 

where b = reading of Baume hydrometer. Water weighs 8.25 
lb. per gal. at 62°. The weight of a gallon of oil at this tempera¬ 
ture is 8.25^ lb. and its heat value is very nearly 

18 650 + 40(6 — 10) B.t.u. per lb. 

Oils are sold by bulk. They expand 0.00055 of the original 
volume per degree rise in temperature. Fuel oil contains C, 
0.80 to 0.87 and H 2 , 0.10 to 0.14, the balance being 0 2 and N 2 
with (frequently) very small traces of sulphur. 

Prob. 317. How much heat is contained in 50 gal. of fuel oil 
of 30° Baume density at 62° F.? (Ans., 7 000 000 B.t.u.) 
What will be the bulk of this oil at 92° F.? (Ans., 50.83 gal.) 

The Diesel Cycle 

160. Ideal Cycle. The ideal Diesel cycle draws in a charge of 
pure air at atmospheric pressure ( ea , Fig. 68) and compresses it 
adiabatically to about 500 lb. pressure. The temperature thus 



220 


Thermodynamics, Abridged 


attained is so high that upon injection of the fuel it immediately 
ignites. The delivery of fuel is so retarded that combustion 
occurs at constant pressure, be, hence there is no explosive 
action. The supply of fuel is cut off and combustion ceases at c. 



Adiabatic expansion follows and the exhaust valve opens at d. 
The pressure drops instantaneously to that of the atmosphere, 
at which pressure the burnt gases are expelled. 

Under these ideal conditions, the efficiency is, 


e ~ lh c - 1 “ Hu 

wl(T d — To) T d -T a 

wk(T c - T„) y(T c - T„) ■ 

The ideal mean effective pressure is (following Art. 151) 

Hh e 

p m = 5.403 77 . ± v - = SAOSBe, 

' a V b 


(92) 


(93) 


where B = heat supplied per cu. ft. of piston displacement. 
For average fuel oil containing C 0.85, H 2 0.12, 0 2 0.01, the vol¬ 
ume of air at standard conditions required for ideal mixture is 
about 200 cu. ft. per lb. of fuel. Hence in the ideal cycle with 
e v = 1.0 (Art. 155), B = 19 500 ^ 200 = 97J, very nearly, 
and p m = 536e. Actual values are less than half this (Art. 162). 










Internal Combustion Engines 


221 


0.63 

<1> 

b m 

§ 0.61 

^ 0.60 

^ 0.59 

(3 

0.58 

0.57 


Prob. 318. Given t a = 40° F., C = p b \ Va = 34, T c = 2 T b , 
y = 1A, find e and p m . (Ans., 0.574, 308 lb.) 

161. Variation of Ideal Efficiency. The Diesel cycle is more 
efficient than the Otto. It is rather unique among heat engine 
cycles, in that its efficiency 

INCREASES AT LIGHT LOADS. 

Fig. 69 plots values of e 
against the ratio of expan¬ 
sion, r = V d /V c = V a l V c , 

Fig. 68. This is based on 
t a = 40° F., C = 34, as in 
Prob. 318, with y = 1.4. 

High ratios of expansion im¬ 
ply low mean effective pres¬ 
sures. Compare the diagrams 
abed and abe'd', Fig. 68. Since 
the ideal value of p m varies 
directly with e (Equation 
(93)) the underload condition 
must be one of (a) poor mix- 


\ 








Ys, 







Vc 

r 







% 







V3 



















<4 

y 


















\ 


120 

1104 

4 

90 | 
80 | 
70 g 

6ol 

to 

50 | 

40 s 
30 


5 6 7 8 9 10 II, 12 

Ratio of Expansion r=Va/Vc 


Fig. 69. Diesel Cycle at Various Loads, 
ture, (6) low volumetric efficiency or (c) low relative efficiency. 


162. The Diesel Engine in Practice. In actual engines, the 
efficiency occasionally, though not frequently, shows an increase 
with decrease of load until the load has been reduced to about 
half its normal value. The heat losses then begin to offset the 
cyclic gain, and below this point the efficiency decreases. In 
most cases, the efficiency falls off with decrease of load over the 
whole range. Most Diesel engines are four-cycle, single-acting. 
The two-cycle arrangement is sometimes used, requiring that 
the air charge be delivered at 20 to 30 lb. pressure. This con¬ 
sumes about 4 per cent, of the engine power. The two-cycle 
Diesel engine gives 70 to 80 per cent, more output than the 
four cycle, but its relative and actual efficiencies are about 10 
per cent. less. Mechanical efficiencies of Diesel engines are 
from 0.70 to 0.75. The high compression used requires that the 


























222 


Thermodynamics, Abridged 


fuel be supplied by a compressed air blast (Art. 159) at about 
1000 lb. pressure. From 16 to 34 cu. ft. of free air (Art. 42) 
are required per brake hp.-hr., and the compressor absorbs 4 to 7 
per cent, of the power of the engine. 

Diagram factors in well-designed engines are around / = 0.50 
to 0.62. The corresponding value of pj, the, actual mean 
effective pressure, is 5.403 e v feB', where B' = B.t.u. per cu. ft. 
of combustible mixture at standard conditions. The value 
of p m ' may also be computed from Fig. 68, 


in which 


Pm = 


Pb( 1 + c) / 71(1 + C — c r) 
n — 1 \ r(l + c) 



(94) 


n = exponent of polytropic curves ab and cd, 
c = clearance = F& -f- (V a — Vb) = 1 -f* ( C 1/n — 1 ), 
r — ratio of expansion = Vd/V C) 

C = compression ratio = pblp a • 

For the usual conditions pb — 500, C = 34, if we take n = y 
= 1.4 and compute c as 0.088, this becomes 1360(1.4/r — 1/r 1,4 ) 
— 114. The curve is plotted in Fig. 69 for n = 1.3 (which 
alters the value of c) and with which 

p m '= 1785 101- 05) 


This last expression may be used with good results in actual 
engines, and there is a satisfactory reason why. It is based 
simply on the shape of the pV diagram, and is subject only to 
such error as may arise from minor irregularities or abnormally 
shaped curves. The cut-off at rated load is usually made to 
occur at about 1/10 stroke, i.e., ( V c — Vb) = 0.10(F o — Vb), 
Fig. 68. With the usual compression, this corresponds with a 
value of r around 6. Fig. 69 shows the corresponding value of 
p m ' to be 109. In ordinary practice the value of p m r is from 100 
to 120 lb. at rated load. 

Prob. 319. Find c and r in Prob. 318. (Ans., c = 0.088, 
r = 6.22.) 





Internal Combustion Engines 


223 


Prob. 320. In Prob. 318, if / = 0.50 and e M = 0.75, what 
weight of oil (19 500 B.t.u. per lb.) will be used per brake hp.-hr.? 
(Ans., 0.61 lb.) 

Prob. 321. In Prob. 318, if B' = 97§ and / = 0.50, what is 
the volumetric efficiency when p m ' = 100? (Ans., e v = 0.66.) 

Prob. 322. If C = 34, n = 1.3, find c and r when cut-off is 
at 10 per cent, of stroke. (Ans., c = 0.071, r = 6.26.) 

Prob. 323. In Prob. 322, find p m ' from Equation (95). (Ans., 
105 lb.) 

Prob. 324. With p m ' as in Prob. 323, find the dimensions of a 
six-cylinder four-cycle single-acting engine to develop 2400 hp. 
at 1000 ft. piston speed and 200 r.p.m. (Ans., diameter, 25.3 in.; 
stroke, 30 in.) 

Tests of Internal Combustion Engines 

163. Methods of Stating Efficiency. If Q = low heat value 
of fuel (Art. 155), B.t.u. per lb., W = weight of fuel used per 
Ihp.-hr., the actual indicated thermal efficiency is 

_ ^ 
e “~ WQ ■ 

If e = efficiency of ideal cycle by Equation (85), the relative 
efficiency is e R = e a -7- e. If e M = mechanical efficiency, the 
brake thermal efficiency is e a e M and the fuel rate per brake 
hp.-hr. is W -T- e M lb. The indicated heat rate is 42.42 -s- e a , 
B.t.u. per Ihp.-min. The brake heat rate is 42.42 -f- e a e M . 
Indicator practice is less satisfactory than with steam engine, 
hence results are often referred to brake hp. Engines using 
producer gas are often reported on the basis of the fuel rate. 
If e P = efficiency of producer and Q P = heat value of producer 
fuel, B.t.u. per lb., the efficiency of the plant from fuel to cylinder 
is e P e a : from fuel to brake, it is e P e a e M . The fuel rate is 2545 
-f- Q P e P e a lb. per Ihp.-hr. The value of e P is around 0.75 in good 
operation. 

Prob. 325. In Probs. 318 and 320, state the actual indicated 
thermal, relative and brake thermal efficiencies, the brake fuel 


224 


Thermodynamics, Abridged 


rate and the indicated and brake heat rates. (Ans., 0.287: 
0.50: 0.215: 0.61 lb.: 148 B.t.u.: 197 B.t.u.) 

Prob. 326. An engine using producer gas, generated at an 
efficiency of e P = 0.72, has an actual indicated thermal efficiency 
of 0.30 and a mechanical efficiency of 0.85. The coal used con¬ 
tains 12 725 B.t.u. per lb. What weight of coal is consumed 
per brake hp.-hr.? (Ans., 1.09 lb.) 

164. Economy Determined by Compression. As indicated by 
theory (Art. 150), the efficiency at full load is chiefly a matter 
of compression. Good four-cycle gasoline and kerosene engines* 
with the low compressions prevailing for those fuels, give values 
of e a from 0.15 to 0.24, the first figure being realized even in 



50 100 150 ZOO 250 300 350 400 450 500 
Approximate CompressioriLb. per 5q.ln. Absolute 
(p^,Figs.64and 68) 


Fig. 70. Economy vs. Compression in Oil Engines. 

very poor installations. With higher compressions, on producer 
gas, natural gas, blast furnace gas and some illuminating gases, 
full load values of e a are from 0.25 to 0.32, occasionally rising to 
nearly 0.40. Fig. 70 illustrates guaranteed performances of 
a number of oil engines, including Diesel engines. A law is clearly 
suggested. Diesel engines attain 0.40 actual efficiency on fuel 
oil or kerosene. Some of the very high-pressure Otto cycle oil 
engines almost reach this figure. 

165. Economy in Relation to Load. Fig. 71 shows a uniform 
type of curve, widely different from that obtained with steam 

















Internal Combustion Engines 


225 


engines or steam turbines (Figs. 29 and 60). In all cases the 
efficiency is a maximum at maximum load. This applies 
whether governing is as in Otto cycle gas engines (Art. 153), 
by the Diesel variable cut-off method, or by varying speed as in 
the case of the Daimler engine. None of the curves for Diesel 
engines show the high efficiency at light load referred to in Art. 
162. 

Prob. 327. In Fig. 70, what is the probable brake fuel rate 
of an engine using 7 atmospheres compression? (Ans., 0.834 lb.) 
If the pressure at the beginning of compression is one atmosphere, 
what is the corresponding ideal efficiency by the dotted curve 



Perce nt of Rated (B rake) h. p. 


Fig. 71. Variation of Economy with Load, Internal Combustion Engines. 

of Fig. 66? At a mechanical efficiency of 0.80, with fuel con¬ 
taining 19 500 B.t.u. per lb., what relative efficiency as com¬ 
pared with Equation (88) is thus implied? (Ans., 0.56.) 

Prob. 328. What weight of 30° fuel oil should be used per 
brake hp.-hr. by the best Diesel engine of Fig. 71, at full load? 
At half load? (Ans., 0.42 lb.: 0.51 lb.) 













































































































































































226 


Thermodynamics, Abridged 


Prob. 329. If the Daimler engine of Fig. 71 used 68° gasoline, 
how many gallons would it require per minute at 70 per cent, of 
its full load? (Ans., 0.0213.) 

166. Heat Balance. The steam engine discharges the largest 
proportion of the heat it receives to the condenser. The gas 
engine divides its heat supply into three (very roughly equal) 
parts: one-third becomes indicated work, one-third is discharged 
at the exhaust, one-third is carried off by the cooling water 
discharged from the jackets. More accurately, indicated work 
accounts for 15 to 46 per cent.: jackets, for 32 to 41 per cent.: 
and exhaust for 24 to 33 per cent.: with a small unaccounted-for 
balance. The proportion to the jackets is low in very large 
engines. 

Jacket water enters at 60° to 80° and emerges at 120° to 170°. 
The former exit temperature must be maintained if the fuel 
contains much hydrogen,in order that preignition maybe avoided. 
Water at these discharge temperatures has very little value 
for heating purposes. The heat is there, but the availability 
is not. If e a = efficiency of engine, j = proportion of heat 
supplied which is carried off by the jacket water, the heat 
removed by the water is 2545j 4- e a , B.t.u. per hp.-hr. This is 
equal to w w (t 2 — h) where w w = weight of jacket water per 
hp.-hr. and ti and t 2 are its inlet and outlet temperatures. The 
water pressure must be sufficient to ensure through circulation. 

The exhaust heat (the amount of which may be computed in a 
similar way) is at high temperature (Art. 154) and is hence 
much more available. The transfer of this heat to water (pre¬ 
ferably the jacket water itself) is the usually attempted method 
of utilizing it. This is done in a device like an economizer 
(Art. 124). The high gas velocity causes much difficulty in 
handling it, and it is a troublesome matter to avoid producing 
serious back pressure at the engine. Some 30 per cent, of the 
heat in the exhaust gases may be recovered under good condi¬ 
tions. If this is recovered as hot water, the best use to make of 
this water is for heating buildings. If the heat is recovered as 
steam, to be used in an engine, the very low efficiency of the 
steam engine enters into the matter. 


Inteknal Combustion Engines 


227 


Prob. 330. How much water per hour is required in the 
jackets of a 1000 hp. engine having an actual efficiency of 0.25 
and a jacket loss of 0.35, if the water temperature rises from 70° 
to 140°? (Ans., 50 900 lb.) 

Prob. 331. The loss to the exhaust in the above engine is 
0.30, of which amount 28 per cent, is recovered. How many 
B.t.u. are recovered per hr.? (Ans., 853 000.) If this is all 
transferred to the jacket water, how much will that water be 
heated above 140°? (Ans., 16.8°, or to 156.8°.) If, on the 
contrary, it is used to generate steam at 100 lb. pressure from 
some of that water, how many B.t.u. must be added to each 
lb. of water? (Ans., 1078.4.) How many lb. of steam will be 
generated per hr.? (Ans., 790.) If this steam is used in a 
turbine which consumes 15.8 lb. of steam per Ihp.-hr., what hp. 
will the turbine develop? (Ans., 50.) What is the per cent, 
saving or gain by this complication? (Ans., 5 per cent.) 


INDEX 


Reference is to Article numbers unless otherwise specified. 


absolute temperature, 11 
absolute zero, 11 
action of engine, 7 
adiabatic, 28, 29, 72, 79 
air, 18 

air compressor, 40-49 

air consumption, air engine, 52, 53 

air cooler, 125 

air engine, 38, 39, 50-55 

air machinery, 38-62 

air refrigeration, 57-62 

air-steam mixture, 92 

alcohol, 153 

ammonia, 112, 113, page 148 
ammonia condenser, 125 
automobile engine, 157 
availability of heat, 36 

back pressure, 88 
desirable, 102 
Baume scale, 159 
binary vapor engine, 103 
blast furnace gas, 153, 155 
boiler, 125, 126 
boiling points of gases, 18 
boiling points of liquids, 63 
Boyle’s law, 9, 20 
brine, 115 
brine cooler, 125 
brine system, 115 
British thermal unit, 3 
bucket friction, 138, 143 

calcium chloride brine, 115 
calorie, 3 

calorimeter, throttling, 135 
carbon dioxide, 65, 112 
carbon dioxide refrigerating machine, 
119-121 


carbon monoxide, 18 
Carnot cycle, 34—37, 88-89 
Centigrade temperature, 2 
Charles’ laws, 10 
clearance, 7, 23, 95, 104 
gas engine, 150 
coal gas, 153, 155 

coefficient of heat transmission, 125 
coke oven gas, 153, 155 
combination turbine, 146 
complete expansion cycle, 80 
compound steam engine, 94, 98, 100, 
102, 106-109 

compound turbine, 143-146 
compressed air, 38-56 
compressed air blast, oil engine, 159 
compressed air plant, 55, 56 
compressed air transmission, 56 
compression, 95, 97, 98 
compression, air engine, 50 
Diesel engine, 160, 162 
internal combustion engine, 164 
compression ratio, 38 

Otto cycle, 150, 153, 159 
compressor, air, 40-49 
compressor, refrigeration, 111 
condensation, 78 
cylinder, 95-100 
condenser, 49, 59, 92, 101, 129 
refrigeration, 111 
condensing engine, 88, 94 
constant dryness curve, 72 
entropy, 79 
heat content, 76 
volume, 75 

convergent nozzle, 136 
cooling in air compressor, 41, 43, 47 
air engine 50, 54 
counter flow, 124 


Index. 


229 


critical pressure, 136 
critical temperature, 65 
Curtis turbine, 144 
cycle, 21, 22, 34 
Diesel, 160, 161 
Ericsson, 62 
ideal vapor, 80 
incomplete expansion, 95 
Otto, 149-151 
Rankine, 80, 95 
for refrigeration, 111 
regenerative, 61, 62 
Stirling, 62 

superheated steam, 84-87 
cylinder condensation, 95-100 
cylinder ratio, 106, 108 

Dalton’s law, 13, 90 
Davis formula, 66 
De Laval turbine, 138-142 
dense air ice machine, 60 
diagram, indicator, 95 
diagram factor, 104 

compound engine, 107, 108 
gas engine, 155 
Diesel engine, 160-162 
difference of specific heats, 15, 16 
disgregation work, 2, 18, 63 
direct expansion, 115 
displacement, 7, 23, 42, 115 
specific, 116 
distiller, 130 
divergent nozzle, 136 
double-acting engine, 7 
double-flow turbine, 146 
double-valve engine, 94 
drop, compound engine, 108 
dry compression, 117 
dryness, effect on efficiency, 88 
dry vapor, 64 
dummy piston, 146 

economizer, 125, 128 
economy, gas engine, 165 
effects of heat, 2 
efficiency, 5, 6 
air engine, 54 


efficiency, air refrigeration, 57, 59 
Carnot cycle, 34 
compressed air plant, 55, 56 
cycle, 22 

dense air ice machine, 60 
Diesel engine, 160, 161 
evaporation, 131 
heat engine, 34, 36 
ideal steam engine, 88 
incomplete expansion cycle, 81 
internal combustion engine, 163- 
165 

Joule cycle, 38 

mechanical (gas engine), 156, 157 
nozzle, 137, 141 
Otto cycle, 150 

variable specific heat, 152 
Rankine cycle, 80, 84-87 
for refrigeration, 114, 115, 118 
regenerative cycle, 62 
relative, 93, 94, 97, 155 
steam cycle, 82 
steam engine, 93-103 
Stirling cycle, 62 
turbine, 141, 142, 147, 148 
two-cycle gas engine, 157 
volumetric, 42, 46, 155, 157 
end thrust, 146 
engine action, 7 
entropy, 67-73 
of liquid, 71 
superheated steam, 83 
entropy of vaporization, 71 
equivalent evaporation, 73 
equivalent heat and mechanical units, 
73 

equivalent mean effective pressure, 
107 

ether, 112 
ethyl chloride, 112 
evaporator, 125, 130, 131 
exhaust loss, gas engine, 166 
exhaust steam, quality of, 101 
expander, 60 

expansion in air engine, 50 
in nozzle, 136 
ratio of, 96, 104 


230 


Index. 


expansion valve, 111, 113 
exponentials, valves of, page 12 
external work, 2 
any path, 29 

graphical representation, 19 
polytropic, 26 
superheating, 83 
vapor adiabatic, 79 
vaporization, 65, 74, 77 

factor, diagram, 104 
factor of evaporation, 73, 83 
Fahrenheit temperature, 2 
feed water heater, 125, 129 
feed water temperature, 73 
first law of thermodynamics, 4 
flow of air, 136 
flow of steam, 133-137 
fluid friction, 156 
fluids for refrigeration, 112 
four-cycle engine, 149-156 
four-valve engine, 94 
free air, 42, 52 
free expansion, 136 
friction, nozzle, 134-137 
turbine, 141, 142 
fuel oil, 153, 159 
furnace efficiency, 126 

gases, composition, 155 
heat value, 155 
gasoline, 153 
gauge pressure, 9 
governing gas engines, 153 
gun, gases in, 32 

heat absorbed, any path, 29 
graphical representation, 33 
polytropic, 27 

heat balance, gas engine, 166 
heat chart, 86, 87 

heat expenditure, forming vapor, 73 
beat of liquid, 63 
heat rate, 82, 93, 163 
heat transfer appliances, 123-132 
heat unit, 3 


heat value, gases, 155 
liquids, 159 

heating of liquid, 63, 68 
helium, 18 
high heat value, 155 
hit-or-miss governing, 153 
horse power, air engine, 51 
air refrigeration, 58, 59 
compressing air, 42, 45 
dense air ice machine, 60 
engine, 7 

impulse turbine, 140 
reaction turbine, 146 
steam boiler, 126 
hot air engine, 61, 62 
humidity, 91 
hydrogen, 18 

ice melting effect, 59 
ideal air engine, 53 

air refrigerating machine, 58-62 
steam rate, 82 
ignition, 153, 159 
imperfect gas, 32 
impulse turbine, 138-142 
incomplete expansion cycle, 81, 95 
indicated horse power, 7 
indicator, 7 

indicator diagram, 7, 95, 149, 154 
indirect refrigeration, 115 
initial pressure, 88, 147 
injector, 134 

intercooler, 44, 47, 48, 125 
intercooler pressure, 45, 48 
internal combustion engine, 149-166 
internal energy, 63, 65, 74 
isodiabatic, 31 
isothermal, 20, 29 

jacket, air compressor, 43, 47 
gas engine, 166 
steam, 97, 98, 104 
Joule cycle, 38, 39, 57 
Joule’s law, 18 

kerosene, 153 

kinetic efficiency, 139, 143 



Index. 


231 


latent heat of fusion, 59 
of vaporization, 64, 65 
law of perfect gas, 12 
load vs. efficiency, turbine, 148 
low heat value, 155 
low-pressure turbine, 147 

marine reaction turbine, 146 
steam engine, 99 
mass-flow theory, 125 
maximum efficiency, heat engine, 
34 

flow, nozzle, 136 
mean B.t.u., 3 

mean effective pressure, 7, 23 
air engine, 53 
Carnot cycle, 37 
dense air ice machine, 60 
Diesel engine, 160, 162 
equivalent simple engine, 107 
Joule cycle, 39 
Otto cycle, 151, 153, 155 
Rankine cycle, 80, 84, 87 
refrigeration, 114, 115, 118 
steam engine, 80, 81, 104, 105 
Stirling cycle, 62 
superheated steam, 105 
two-stage air compressor, 45 
mean temperature difference, 124 
mechanical efficiency, 6 
Diesel engine, 162 
gas engine, 156, 157 
refrigeration, 115 
turbine, 142 

mechanical equivalent of heat, 4 
methane, 18 
mixtures, 123 
air and steam, 90-92 
gases, 13 
moist air, 90-92 
molecular weights, 18 
Mollier diagram, 86, 87 
multiple effect evaporator, 131 
multiple expansion engine, 100, 110 
multi-stage air compressor, 48 
refrigeration, 117 
turbine, 143-146 
16 


natural gas, 153, 155 
negative heat of liquid, 63 
negative loop, 3 
negative specific heat, 14 
nitrogen, 18 

non-condensing engine, 94 
nozzle, 136, 137 
nozzle angle, 138 

oil engine, 159 
oil gas, 155 
Otto cycle, 149-151 
outside compression, 157 
overload capacity, 7, 156 
oxygen, 18 

parallel flow, 124 

Parsons turbine, 145, 146 

parts of engine, 7 

perfect gas, 8-37 , 

peripheral speed, 139, 140, 144, 146 

permanent gas, 8-37 

pipe covering, 132 

piston pressures, 106 

piston speed, 7, 104, 115 

polytropic, 24-27, 29 

power-speed curve, marine engine, 99 

preheat, 54, 55 

preheater, 54 

pressure, 8 

pressure-compounded turbine, 144 
pressure-temperature curve, 63 
producer gas, 153, 155 
properties of gases, 18 
of steam, 66, 74 
propulsion, power for, 6 

quadruple-expansion engine, 100,103, 
110 

quality governing, 153 
quality of steam, 135 
quantity governing, 153 

radiant heat, 126 

radiation from steam pipes, 132 

radiator, 125 

Rankine cycle, 80, 84-87, 95 

for refrigeration, 111, 114, 118 


232 


Index. 


ratio of expansion, 88, 96, 104, 161, 
162 

ratio of specific heats, 17 
reaction turbine, 145, 146 
receiver, 106, 125 
reduction gear, 142 
refrigeration, 111-122 
refrigeration, air, 57-62 
Carnot cycle, 35 
coils, 125 

regenerative engine, 61, 62, 103 
reheater, 100 

relative efficiency, 53, 55, 93, 94, 97, 
142, 155 

relative humidity, 91 
relative velocity, 138 
releasing valve gear, 94 
rotation loss, 141 

saturated air, 90 
saturated vapor, 64 
scavenging, 157 

second law of thermodynamics, 36 
simple steam engine, 104, 105 
simple impulse turbine, 138-142 
single-acting engine, 7 
single-stage turbine, 138-142 
single-valve engine, 94 
specific displacement, 116 
specific gravity, liquid fuels, 159 
specific heat, 3 
any path, 29 
gases, 14, 15 
polytropic, 27 
variable, 152 
speed, piston, 104 
speed-power curve, 99 
speed, steam engine, 99, 104 
turbine, 140 
steam, 63-92 

steam consumption, engine, 94, 96 
steam engine, action, 95 

compound, 94, 98, 100, 102 
condensing, 94 
efficiency, 93-103 
heat rate, 93 

mechanical efficiency, 100 


steam engine, non-condensing, 94 
piston speed, 104 
power, 93-100 
quadruple, 100, 103 
regenerative, 103 
triple, 94, 100 
Uniflow, 103 
vacuum, 102 
valves, 94 

steam jacket, 97, 98, 104 
steam power plant, 80 
steam rate, 82, 142 
superheated, 94, 98 
table, 66 
turbine, 133-148 
straight fine law, boiler, 126 
sulphur dioxide, 112 

refrigerating machine, 122 
supercompression, 155 
superheat, 83, 88, 109, 110 
in refrigeration, 117, 118 
superheater, 125, 127 
superheated steam, efficiency, 94, 98, 
148 

mean effective pressure, 105 
superheated vapor, 64, 83-87 
superheating, 72 
supersaturated air, 90 
surface condenser, 125, 129 
surface efficiency, 126 

temperature, 2, 3 

temperature-entropy diagram, 67-73 
temperature ranges, compound en¬ 
gine, 106 

terminal drop, 108 
thermal efficiency, 6 
thermodynamics, definition, 1 
thermodynamic surface, 12 
throttling calorimeter, 135 
timed injection of oil, 159 
tonnage, 59, 115, 116 
total heat of vapor, 65 
total heat-entropy diagram, 86, 87 
total heat-pressure diagram, 87 
transfer of heat, 123-132 
triple-expansion engine, 94, 100, 110 


Index. 


233 


turbine, 133-148 

two-cycle engine, 157, 158, 159, 162 
two-stage air compressor, 44-48 

Uniflow engine, 103 

vacuum, 92, 148 
best for engine, 102 
pump, 49 

valve, steam engine, 94 
valveless gas engine, 157 
Van der Waals equation, 32 
vapor, 8, 63-92 
for heat engine, 89 
refrigeration, 111-122 
vaporization, 64, 65, 72, 74 
variable specific heat, 152 
velocity, adiabatic flow, 133 
velocity compounding, 143, 144 


velocity diagram, 138, 144, 145 
energy, 133 
nozzles, 136, 137 
steam, in reaction turbine, 146 
volume of gas, 8, 12 
liquid, 65, 77 
superheated steam, 83 
vapor, 64, 74, 75 

volumetric efficiency, 42, 46,115, 155, 
157 

water gas, 153, 155 
water required air compressor, 43 
Westinghouse-Parsons turbine, 146 
wet compression, 117 
wet vapor, 64, 72, 74-76 
wire-drawing, 134-137 
work of turbine buckets, 138, 142, 
143, 145, 146 











































































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